# Clarification on the countable additivity of Lebesgue measure

As emphasized in remark 1.2.4, p. 19, of Terence Tao’s book “An introduction to measure theory”, finite additivity doesn’t hold for Lebesgue outer measure $m^{*}(\cdot)$ in general, and therefore it doesn’t hold for Lebesgue measure $m(\cdot)$ either. So, the Lebesgue outer measure $m^{*}(E\cup F)$ of the union of two disjoint sets $E, F$ in the Euclidean metric space $(\mathbb{R}^d, |\cdot|)$ does not necessarily satisfy $m^{*}(E\cup F)=m^{*}(E)+m^{*}(F)$.

If $E,F$ are both Lebesgue measurable, then it holds that $m^{*}(E\cup F)=m^{*}(E)+m^{*}(F)$. Moreover, $m^{*}(E)=m(E)$$m^{*}(F)=m(F)$, which means that the union $E\cup F$ is also Lebesgue measurable and ultimately finite additivity follows as $m(E\cup F)=m(E)+m(F)$.

The main point is that if $E\cup F$ is Lebesgue measurable and $E\cap F=\emptyset$, then finite additivity doesn’t follow. Instead, the disjoint set assumption $E\cap F=\emptyset$ needs to be replaced by the positive distance assumption $\mbox{dist}(E, F)=\inf\{|x-y|: x\in E, y\in F\}>0$ to ensure finite additivity for the outer measure, as explained and proved in lemma 1.2.5, p. 19, of Tao’s book.

Considering this limitation in the applicability of finite additivity, a reader may feel alarmed when reading lemma 1.9.c, p. 21, of David Williams’ book “Probability with martingales”, which states that if $m(S)<\infty$ in a measure space $(S, \Sigma, m)$, then $m(F\cup E) = m(F)+m(E)-m(F\cap E)$ for $F, E\in\Sigma$. If $F\cap E=\emptyset$, then $m(F\cup E) = m(F)+m(E)$, so finite additivity holds.

Recall that a measure $m:\Sigma\rightarrow[0,\infty]$ on $(S,\Sigma)$ has the whole σ-algebra $\Sigma$ on $S$ as its domain. This implies that for any $F, E\in\Sigma$ the measures $m(F),m(E)$ exist by assumption, therefore there is no conflict between the aforementioned statements on finite additivity of measure found in Tao’s and Williams’ book.

It becomes clear that operating on a measure space $(\mathbb{R}^d,\Sigma, m)$ seems to avoid the trouble of having to prove the existence of Lebesgue measure for every set of interest, simply because by definition $m(F)$ exists for every $F\in\Sigma$. However, there is no free lunch. A follow-up question arises, as one then needs to show that $F$ belongs to the defined σ-algebra $\Sigma$, which is not always trivial.

For instance, the experiment of tossing a coin infinitely often is presented in p. 24 of Williams’ book by introducing an associated sample space $\Omega$ and subsequently a probability triple $(\Omega,\mathcal{F},P)$. An event $F$ of possible interest is that the ratio of heads in $n$ tosses tends to $1/2$ as $n\rightarrow\infty$. Even for such a seemingly simple experiment, proving that this event $F$, seen as a set, belongs to the σ-algebra $\mathcal{F}$ on $\Omega$ is already not so trivial to prove.

On a final note, lemma 1.2.15, p. 30, in Tao’s book, proves countable additivity for disjoint Lebesgue measurable sets, which subsumes finite additivity. The proof is easy to follow; the claim is proved first for compact, then for bounded and then for unbounded sets. To conclude the present post, a clarification is made in the proof of the bounded case. In particular, it will be explained why for a bounded Lebesgue measurable set $E_n$ there exists a compact set $K_n$ such that $m(E_n) \le m(K_n)+\epsilon/2^n$.

Since $E_n$ is Lebesgue measurable, it follows from exercise 1.2.7.iv in Tao’s book that for any $\epsilon > 0$ there exists a closed set $K_n$ such that $K_n\subseteq E_n$ and $m^{*}(E_n\setminus K_n) \le \epsilon/2^n$. By applying the countable subadditivity of outer measure (see exercise 1.2.3.iii in Tao’s book), $E_n=K_n\cup (E_n\setminus K_n)$ leads to $m^{*}(E_n)\le m^{*}(K_n)+m^{*}(E_n\setminus K_n) \le m^{*}(K_n)+\epsilon/2^n$. Furthermore, $K_n$ is Lebesgue measurable since it is closed (see lemma 1.2.13.ii in Tao’s book). Thus, Lebesgue outer measure can be replaced by Lebesgue measure to give $m(E_n) \le m(K_n)+\epsilon/2^n$. Finally, $K_n$ is bounded, as a subset of the bounded set $E_n$, and closed, so $K_n$ is compact, which completes the proof.

# Proof of Baire’s category theorem

The book “Probability with martingales” by David Williams provides a proof of Baire’s category theorem in p. 203-204. This blog post reproduces the proof in greater detail to clarify some of the statements that are left to the book reader to confirm.

Baire’s category theorem is stated as follows:

If a complete metric space $S$ can be written as a union $S=\underset{n\in\mathbb{N}}{\cup}F_n$ of a countable sequence of closed sets $F_n,n\in\mathbb{N}$, then at least one of these closed sets contains an open ball.

The theorem was proved for $\mathbb{R}^d$ by René-Louis Baire in his PhD thesis “Sur les fonctions de variable réelles“.

Assume a complete metric space $(S,\rho)$ with some metric $\rho:S\times S\rightarrow \mathbb{R}$, which need not be necessarily the Euclidean metric space $(\mathbb{R}^d, |\cdot|)$.

Start by assuming that none of $F_n,n\in\mathbb{N}$, contains an open ball. Moreover, assume that none of the complements $F_n^c,n\in\mathbb{N}$, is empty. If there exists an empty complement $F_n^c=\emptyset$, then $F_n=S$, which is a trivial case of finite union that will be treated at the end of the proof separately.

Since $F_1$ is closed, $F_1^c$ is open, i.e. $F_1^c=(F_1^c)^{\mathrm{o}}$. Due to $F_1^c$ being non-empty, there exists $x_1\in F_1^c=(F_1^c)^{\mathrm{o}}$. Hence, there exists $\epsilon_1 >0$ so that the open ball $B(x_1,\epsilon_1)$ with center $x_1$ and radius $\epsilon_1$ satisfies $B(x_1,\epsilon_1)\subseteq F_1^c$.

By assumption, $F_2$ contains no open ball, so $B(x_1, \epsilon_1/2)\not\subseteq F_2$, which means that $B(x_1, \epsilon_1/2)\cap F_2^c \neq\emptyset$. Furthermore, $B(x_1, \epsilon_1/2)\cap F_2^c$ is an open set as the intersection of the open sets $B(x_1, \epsilon_1/2)$ and $F_2^c$. Thus, there exist $x_2\in B(x_1, \epsilon_1/2)\cap F_2^c$ and $\epsilon_2>0$ with $\epsilon_2<\min \{\epsilon_1/2,1/2\}$ such that $B(x_2,\epsilon_2)\subseteq B(x_1,\epsilon_1/2)\cap F_2^c$. Note that $B(x_2,\epsilon_2)\subseteq B(x_1,\epsilon_1/2)$.

Inductively, choose $\epsilon_n<\min \{\epsilon_{n-1}/2,1/n\},n\in\mathbb{N}$, to introduce the n-th ball $B(x_n,\epsilon_n)\subseteq B(x_{n-1},\epsilon_{n-1}/2)\cap F_n^c$. This way, a countable sequence of nested balls is defined with $B(x_n,\epsilon_n)\subseteq B(x_{n-1},\epsilon_{n-1}/2)$. It will be shown that the centers of these balls make up a Cauchy sequence $(x_n)$. According to the Archimedean property, for any $\epsilon > 0$, there exists $n_0\in\mathbb{N}$ such that $0< 1/n_o < \epsilon$. Let $m,n\in\mathbb{N}$ be natural numbers with $m>n_o$ and $n>n_o$. Using the triangle inequality, $\rho (x_m, x_n) \le \rho (x_m, x_{n_o})+\rho (x_{n_o}, x_n)$. Obviously, $x_n\in B(x_n,\epsilon_n)$. Since $n>n_o$, it follows that $B(x_n,\epsilon_n)\subseteq B(x_{n_o}, \epsilon_{n_o}/2)$, so $x_n\in B(x_{n_o}, \epsilon_{n_o}/2)$. Apparently, $x_n\in B(x_{n_o}, \epsilon_{n_o}/2)$ implies $\rho(x_n,x_{n_o})<\epsilon_{n_o}/2$. As $\epsilon_{n_o}<1/n_o$ and $1/n_o < \epsilon$, it follows that $\rho(x_n,x_{n_o})<\epsilon/2$. Similarly, $\rho(x_m,x_{n_o})<\epsilon/2$, which leads to $\rho (x_m, x_n) < \epsilon$, so the ball centers constitute the Cauchy sequence $(x_n)$. The metric space $(S, \rho)$ has been assumed to be complete, so the Cauchy sequence $(x_n)$ is convergent, therefore its limit $x:=\lim{x_n}$ exists in $S$.

The next step is to show that the limit $x$ of the ball centers $(x_n)$ belongs to the intersection of these balls, i.e. $x\in\underset{n\in\mathbb{N}}{\cap}B(x_n,\epsilon_n)$. Let $n\in \mathbb{N}$ and $m\in\mathbb{N}$ with $m>n$. By applying the triangle inequality, $\rho (x, x_n) \le \rho (x, x_m)+\rho (x_m, x_n)$. Notice that $m>n$ yields $x_m\in B(x_m,\epsilon_m)\subseteq B(x_n, \epsilon_n/2)$, hence $\rho (x_m, x_n) <\epsilon_n/2$, which leads to $\rho (x, x_n) \le \rho (x, x_m)+\epsilon_n/2$. Observe that $(x_m)$ is a subsequence of $(x_n)$, which means that $\underset{m\rightarrow\infty}{\lim}\rho (x, x_m)=0$. So, taking the limit as $m\rightarrow \infty$ produces  $\rho (x, x_n) \le \epsilon_n/2$, whence $x\in B(x_n,\epsilon_n)$ for any $n \in \mathbb{N}$, thus $x\in\underset{n\in\mathbb{N}}{\cap}B(x_n,\epsilon_n)$.

It is obvious from $B(x_n,\epsilon_n)\subseteq B(x_{n-1},\epsilon_{n-1}/2)\cap F_n^c$ that $B(x_n,\epsilon_n)\subseteq F_n^c$. Consequently, $x\in\underset{n\in\mathbb{N}}{\cap}B(x_n,\epsilon_n)\subseteq\underset{n\in\mathbb{N}}{\cap}F_n^c=\left(\underset{n\in\mathbb{N}}{\cup}F_n\right)^c=S^c$. The contradiction $x\in S^c=\emptyset$ has been reached, which completes the proof.

The case of finite union $S=\overset{k}{\underset{n=1}{\cup}}F_n$ is simpler, as it is not needed to construct an infinite sequence of open balls and then take the limit of their centers. Instead, it suffices to observe that the center $x_k$ of the smallest ball $B(x_k,\epsilon_k)$ satisfies $x_k\in\overset{k}{\underset{n=1}{\cap}}B(x_n,\epsilon_n)\subseteq\overset{k}{\underset{n=1}{\cap}} F_n^c$, leading to the analogous contradiction $x_k\in S^c=\emptyset$.

On a final note, it is emphasized that the existence of the countable sequence of ball centers $(x_n)$ is based on the axiom of choice.

# Clarification on the proof that an open set is a countable union of almost disjoint boxes

To establish that the Lebesgue outer measure of any open set in the Euclidean metric space of $\mathbb{R}^d$ is equal to the volume of any partitioning of that set into almost disjoint boxes, lemma 1.2.11 in p. 24 of the book “An introduction to measure theory” by Terence Tao first states that any open set $E\subseteq \mathbb{R}^d$ can be expressed as a countable union of almost disjoint boxes (and in fact as a countable union of almost disjoint closed cubes).

Note that this lemma is a generalization of the fact that every open subset of $\mathbb{R}$ can be expressed as a countable union of disjoint open intervals (see theorem 1.3, p. 6, in the book “Real analysis: measure theory, integration and Hilbert spaces” by Elias M. Stein and Rami Shakarchi and also lemma 2 of this blog post of mine).

The purpose of the present blog post is to elaborate on a detail in the proof of lemma 1.2.11 of Tao’s book. In particular, it will be explained why  every closed dyadic cube $Q$ is contained in exactly one maximal cube $Q^{*}$, without reproducing the rest of the proof.

Introduce the notation $Q_{n,\mathbf{i}}=\overset{d}{\underset{k=1}{\times}}\left[\frac{i_k}{2^n},\frac{i_k+1}{2^n}\right]$ to indicate the dependence of closed dyadic cubes on $n\in\mathbb{N}\cup \left\{0\right\}$ and $\mathbf{i}=(i_1,i_2,\cdots, i_d)\in\mathbb{Z}^{d}$. For every $\mathbf{i}=(i_1,i_2,\cdots, i_d)\in\mathbb{Z}^{d}$ for which there exists a closed dyadic cube $Q_{n,\mathbf{i}}$ contained in $E$, choose the biggest closed dyadic cube $Q_{n,\mathbf{i}}$ in $E$, i.e. choose $n_0=\underset{n}{\min}\left\{n\in\mathbb{N}\cup \left\{0\right\}:Q_{n,\mathbf{i}}\in E\right\}$. It is now obvious that by having capped the cubes by a sidelength of 1, there exists a maximum cube $Q_{n_0,\mathbf{i}}$ among $\left\{Q_{n,\mathbf{i}}\in E:n\in\mathbb{N}\cup \left\{0\right\}\right\}$, which is a maximal cube among those contained in $E$.

What is left to show is that every closed dyadic cube $Q$ can’t be contained in two or more maximal cubes. Assume that $Q$ is contained in two distinct maximal cubes $Q^*$ and $Q^{**}$, i.e. $Q\subseteq Q^*$ and $Q\subseteq Q^{**}$ with $Q^*\neq Q^{**}$. The dyadic nesting property then leads to the contradiction $Q^*\subseteq Q^{**}$ or $Q^{**}\subseteq Q^*$ for the maximal cubes $Q^*$ and $Q^{**}$, since $Q\subseteq Q^*$ and $Q\subseteq Q^{**}$ excludes the possibility of $Q^*$ and $Q^{**}$ being almost disjoint.

# Clarification on the proof of Carathéodory’s lemma

Carathéodory’s lemma, as it appears in p. 197 (appendix A1) of the book “Probability with martingales” by David Williams, is stated as follows:

Let $\lambda$ be an outer measure on the measurable space $(S,\mathcal{G})$. Then the λ-sets in $\mathcal{G}$ form a σ-algebra $\mathcal{L}$ on which $\lambda$ is countably additive, so that $(S,\mathcal{L},\lambda)$ is a measure space.

Three aspects of the proof of Carathéodory’s lemma provided in Williams’ book are clarified in this blog post.

Definition of λ-system

The concept of λ-system, which is used implicitly but it is not defined in Williams’ book, is introduced in this post.

A collection $\mathcal{C}$ of subsets of a set $S$ is called a λ-system on $S$ if

1. $S\in\mathcal{C}$,
2. $L\in\mathcal{C}\Rightarrow L^{c}\in\mathcal{C}$ (it is closed under complements),
3. $(\forall n\in\mathbb{N})L_n\in\mathcal{C}$ with $L_i\cap L_j=\emptyset$ for $i\neq j$ it holds that $\underset{n\in\mathbb{N}}{\cup}L_n\in\mathcal{C}$ (it is closed under countable disjoint unions).

Not that the only difference between a λ-system and a σ-algebra is that the former is closed under countable disjoint unions while the latter is closed under countable unions. Moreover, the first condition on the definition of a λ-system could be alternatively set to $\emptyset\in\mathcal{C}$ instead of $S\in\mathcal{C}$ due to closure under complementarity, i.e. due to the second condition of the definition.

Lemma

If a collection of subsets of a set $S$ is a λ-system and a π-system on $S$, it is also a σ-algebra on $S$.

This lemma is used without being proved in Williams’ book for proving Carathéodory’s lemma. In what follows, the lemma will be proved before proceeding with the proof of Carathéodory’s lemma.

Although not relevant to subsequent developments, it is mentioned that a σ-algebra on a set $S$ is also a λ-system on $S$ as it can be trivially seen from the involved definitions.

Proof of the lemma

Let $\mathcal{C}$ be a collection of subsets $S$ that is both a λ-system and a π-system on $S$. To show that $\mathcal{C}$ is a σ-algebra on $S$, it suffices that it is closed under countable unions.

Let $B_n\in\mathcal{C},n\in\mathbb{N}$. The main idea is to express the collection $\{B_n: n\in\mathbb{N}\}$ as a collection $\{L_n: n\in\mathbb{N}\}$ of pairwise disjoint sets ($L_i\cap L_j=\emptyset$ for $i\neq j$) so that $\underset{n\in\mathbb{N}}{\cup}B_n=\underset{n\in\mathbb{N}}{\cup}L_n$. Along these lines, define $L_n:=B_n\setminus \left(\overset{n-1}{\underset{k=1}{\cup}} B_k\right)$.

Obviously, $\underset{n\in\mathbb{N}}{\cup}L_n\subseteq \underset{n\in\mathbb{N}}{\cup}B_n$. To prove the converse set inequality, let $x\in\underset{n\in\mathbb{N}}{\cup}B_n$ and assume that $(\forall n\in\mathbb{N}) x\not\in B_n\setminus \left(\overset{n-1}{\underset{k=1}{\cup}} B_k\right)$. In this case, for each $n\in\mathbb{N}$, either $x\not\in B_n$ or $x\in B_n\cap\left(\overset{n-1}{\underset{k=1}{\cup}} B_k\right)$. There is at least one $n_{*}\in\mathbb{N}$ such that $x\in B_{n_{*}}\cap\left(\overset{n_{*}-1}{\underset{k=1}{\cup}} B_k\right)$, otherwise $(\forall n\in\mathbb{N}) x\not\in B_n$ leads to the contradiction $x\not\in\underset{n\in\mathbb{N}}{\cup}B_n$. Let $n_o\in\mathbb{N}$ be the minimum natural for which $x\in B_{n_{o}}\cap\left(\overset{n_{o}-1}{\underset{k=1}{\cup}} B_k\right)$. In turn, $x\in \overset{n_{o}-1}{\underset{k=1}{\cup}} B_k\Rightarrow (\exists i. Due to $n_o$ being the smallest natural for which $x\in B_{n_{o}}\cap\left(\overset{n_{o}-1}{\underset{k=1}{\cup}} B_k\right)$, it is deduced that $x\not\in B_i\cap\left(\overset{i-1}{\underset{k=1}{\cup}} B_k\right)$, hence $x\not\in\left(\overset{i-1}{\underset{k=1}{\cup}} B_k\right)$. Thus, $(\exists i\in\mathbb{N}) x\in B_i\setminus \left(\overset{i-1}{\underset{k=1}{\cup}} B_k\right)$, which is a contradiction. Thereby, $\underset{n\in\mathbb{N}}{\cup}B_n\subseteq \underset{n\in\mathbb{N}}{\cup}L_n$, and this establishes the equality $\underset{n\in\mathbb{N}}{\cup}L_n= \underset{n\in\mathbb{N}}{\cup}B_n$.

Assume that there are $i,j\in\mathbb{N}$ with $i\neq j$ and $L_i \cap L_j \neq \emptyset$. Let $x\in L_i \cap L_j$. Without loss of generality assume that $i. Then $x\in B_i$ with $i, while $x\in L_j=B_j\setminus \left(\overset{j-1}{\underset{k=1}{\cup}} B_k\right)$, which means that $(\forall k\in\mathbb{N})$ with $k it holds that $x\not\in B_k$, so a contradiction has been reached. Thereby, the sets $L_n, n\in\mathbb{N}$, are pairwise disjoint.

It has thus been shown that the collection $L_n:=B_n\setminus \left(\overset{n-1}{\underset{k=1}{\cup}} B_k\right)$ consists of pairwise disjoints sets that satisfy $\underset{n\in\mathbb{N}}{\cup}L_n=\underset{n\in\mathbb{N}}{\cup}B_n$.

Notice that $L_n:=B_n\setminus \left(\overset{n-1}{\underset{k=1}{\cup}} B_k\right)=B_n \overset{n-1}{\underset{k=1}{\cap}}B_k^c$. Since $\mathcal{C}$ is a λ-system, $B_k^c\in\mathcal{C}$ for the various $k$. Moreover, $\mathcal{C}$ is a π-system, hence the finite intersection $L_n=B_n \overset{n-1}{\underset{k=1}{\cap}}B_k^c$ is also in $\mathcal{C}$. Since the collection $L_n:=B_n\setminus \left(\overset{n-1}{\underset{k=1}{\cup}} B_k\right)$ is a disjoint union of elements $L_n\in\mathcal{C}$ and $\mathcal{C}$ is a λ-system, it follows that the union $\underset{n\in\mathbb{N}}{\cup}L_n= \underset{n\in\mathbb{N}}{\cup}B_n$ is also in $\mathcal{C}$.

Since the countable (but not necessarily disjoint) union $\underset{n\in\mathbb{N}}{\cup}B_n$ of any collection $\{B_n: n\in\mathbb{N}\}$ of sets $B_n\in\mathcal{C}$ is also in $\mathcal{C}$, it follows that $\mathcal{C}$ is a σ-algebra.

First clarification

The above lemma explains why the proof of Carathéodory’s lemma in Williams’ book states that it suffices to show that for a countable collection $\{L_n: n\in\mathbb{N}\}$ of disjoint sets $L_n\in\mathcal{L}$ it holds that $\underset{n\in\mathbb{N}}{\cup}L_n\in\mathcal{L}$. The conclusion then extends to any such countable union of sets, disjoint or not.

Second clarification

It is mentioned in p. 197 of Williams’ book that from

$\lambda (G)\ge\overset{n}{\underset{k=1}{\sum}}\lambda(L_k\cap G)+\lambda(L^c\cap G)$

follows

$\lambda (G)\ge\overset{\infty}{\underset{k=1}{\sum}}\lambda(L_k\cap G)+\lambda(L^c\cap G)$.

To see why this is the case, recall that the outer measure $\lambda:\mathcal{G}\rightarrow [ 0, \infty ]$ takes in values in $\lambda (G)\in[0, \infty]$ for any $G\in\mathcal{G}$.

Distinguish two cases. If $\lambda (G)\in [0,\infty )$ (i.e. if $\lambda (G)$ is finite), then the sequence $a_n=\overset{n}{\underset{k=1}{\sum}}\lambda(L_k\cap G),n\in\mathbb{N}$, is a bounded increasing sequence, therefore it converges, which means that the limit $\overset{\infty}{\underset{k=1}{\sum}}\lambda(L_k\cap G)=\underset{n\rightarrow\infty}{\lim}a_n < \infty$ exists, so taking limits leads from the former to the latter inequality in the book.

If $\lambda (G)=\infty$, then $\infty \ge\overset{\infty}{\underset{k=1}{\sum}}\lambda(L_k\cap G)+\lambda(L^c\cap G)$ holds trivially.

Third clarification

To show that $\lambda (L)=\underset{k\in\mathbb{N}}{\sum}\lambda (L_k)$ for $L=\underset{k\in\mathbb{N}}{\cup}L_k$, notice first that

$\lambda (L)\le\underset{k\in\mathbb{N}}{\sum}\lambda (L_k)$

follows from the countable subadditivity of the outer measure $\lambda$.

Moreover, setting $G=L$ in equation (d) of p. 197 gives

$\lambda (L)\ge\underset{k\in\mathbb{N}}{\sum}\lambda (L_k\cap L)+\lambda (L^c\cap L)=\underset{k\in\mathbb{N}}{\sum}\lambda (L_k)$,

which concludes the argument.

# Proof: a representation of the Borel σ-algebra on reals

This post elaborates on the proof of a proposition in page 17 of the book “Probability with martingales” by David Williams. The relevant proposition is first stated; the Borel σ-algebra $\mathcal{B}:=\mathcal{B}(\mathbb{R})$ on the set $\mathbb{R}$ of real numbers is given by $\mathcal{B}=\sigma (\pi (\mathbb{R}))$, where $\sigma(\pi(\mathbb{R}))$ denotes the sigma algebra on $\mathbb{R}$ generated by $\pi (\mathbb{R})=\{(-\infty, x]:x\in\mathbb{R}\}$.

It is reminded that the Borel σ-algebra $\mathcal{B}(S)$ on a set $S$ is the σ-algebra $\sigma(\mathcal{O})$ generated by the family $\mathcal{O}$ of open subsets of $S$, i.e. $\mathcal{B}(S):=\sigma(\mathcal{O})$.

Two lemmas are provided before proceeding with the proof.

Lemma 1

Let $S$ be a set, and let $\mathcal{C}_1,\mathcal{C}_2$ be two collections of subsets of $S$. If $\mathcal{C}_1\subseteq\sigma(\mathcal{C}_2)$, then $\sigma(\mathcal{C}_1)\subseteq\sigma(\mathcal{C}_2)$.

Proof of lemma 1

$\sigma(\mathcal{C}_2)$ is a σ-algebra on $S$ that contains $\mathcal{C}_1$ and $\sigma(\mathcal{C}_1)$ is the intersection of all σ-algebras on $S$ that contain $\mathcal{C}_1$, so $\sigma(\mathcal{C}_1)\subseteq\sigma(\mathcal{C}_2)$.

Lemma 2

Every open subset $G$ of $\mathbb{R}$ in the Euclidean metric space $(\mathbb{R},\|\cdot\|)$ can be represented uniquely as a countable union of disjoint open intervals.

Proof of lemma 2

This lemma appears also as theorem 1.3, p. 6, in the book “Real analysis: measure theory, integration and Hilbert spaces” by Elias M. Stein and Rami Shakarchi, where a partial proof is provided as uniqueness is not established. In what follows, the proof from Stein’s and Shakarchi’s book is reproduced in more detail, plus uniqueness is proved.

Let $G$ be an open subset of $\mathcal{R}$ and $x\in G$. Introduce the sets $\mathcal{A}_x:=\{a\in\mathbb{R}: a and $\mathcal{B}_x:=\{b\in\mathbb{R}: x. Moreover, let $a_x:=\mbox{inf}\mathcal{A}_x$ and $b_x:=\mbox{sup}\mathcal{B}_x$, allowing for $a_x,b_x$ to be infinite. Consider also the interval $I_x:=(a_x,b_x)$.

It will be shown that $I_x\subseteq G$. Let $y\in I_x$. Assume that $(\forall a\in\mathcal{A}_x)y\le a$, in which case $y$ is a lower bound of $\mathcal{A}_x$, so $y\le a_x$, thus contradicting the assumption $y\in I_x=(a_x, b_x)$. Hence, there exists $a\in\mathcal{A}_x$ such that $a. By a symmetric argument, there exists $b\in\mathcal{B}_x$ such that $y. So, $y\in (a,b)\subseteq (a_x,b_x)$. If $y=x$, then obviously $y\in G$. If $y\neq x$, then $y\in (a,x)\cup (x, b)$. Note that $a\in\mathcal{A}_x$ means that $(a,x)\subseteq G$. So, if $y\in (a,x)$, then $y\in G$. Similarly, if $y\in (x,b)$, then $y\in G$. Thus, $y\in I_x\Rightarrow y\in G$, which means that $I_x\subseteq G$.

$(\forall x\in G) I_x\subseteq G\Rightarrow\underset{x\in G}{\bigcup}I_x\subseteq G$. Furthermore, let $y\in G$. As $G=G^{o}$, there exists $r>0$ such that $(x-r,x+r)\subseteq G$. It follows from $y-r and $(y-r,y)\subseteq (y-r,y+r)\subseteq G$ that $y\in\mathcal{A}_y$. In a similar way, $y\in\mathcal{B}_y$. So, $a_y\le y\le b_y\Rightarrow y\in I_y\subseteq\underset{x\in G}{\bigcup}I_x$. Hence, $G=\underset{x\in G}{\bigcup}I_x$.

Assume that $I_x\cap I_y\neq\emptyset$, in which case $I_x\cup I_y$ is also an open interval. It is straightforward to confirm that if $I$ is an open interval containing $x$ and being contained in $G$, then $I\subseteq I_x$. So, $I_x\cup I_y\subseteq I_x$ and $I_x\cup I_y\subseteq I_y$, which yields $I_x=I_y$. It has been shown that $I_x\cap I_y\neq\emptyset\Rightarrow I_x=I_y$. By contraposition, $I_x\neq I_y\Rightarrow I_x\cap I_y=\emptyset$, which means that the sets of the collection $\{I_x:x\in G\}$ are pairwise disjoint.

For every set $I_x$ in $\{I_x :x\in G\}$, pick a rational $q_x\in I_x$. The set $S:=\{q_x: q_x\in I_x\}$ of selected rationals is obviously a subset of the set $\mathbb{Q}$ of rationals. $S\subseteq\mathbb{Q}$ implies the cardinality inequality $|S|\le |Q|$, so $S$ is a countable set. Since  $|\{I_x :x\in G\}|=|S|$, it is deduced that the collection $\{I_x :x\in G\}$ is also countable.

It has been proved so far that $G$ can be expressed as a countable union of disjoint open intervals. Although not pertinent to the subsequent proof, it will be shown that such representation is unique for the sake of completeness of exposition.

Let $\{I_k : k\in K\}$ and $\{J_{\ell} : \ell\in L\}$ be two collections of disjoint open intervals such that $G=\underset{k\in K}{\bigcup}I_k=\underset{\ell\in L}{\bigcup}J_{\ell}$.

For any $k\in K$, it holds that $I_k=\underset{\ell\in L}\bigcup(I_k\cap J_{\ell})$. Moreover, the sets of the collection $\{I_k\cap J_{\ell}:\ell\in L\}$ are pairwise disjoint. Since $I_k$ and $J_{_\ell}$ are open, $(I_k\cap J_{\ell})^{o}=I_k^{o}\cap J_{\ell}^{o}=I_k\cap J_{\ell}$, which means that $I_k\cap J_{\ell}$ is also open.

In the collection $\{I_k\cap J_{\ell}:\ell\in L\}$, at least one set is non-empty, otherwise trivially $G=\emptyset$.

In the Euclidean metric space $(\mathbb{R},\|\cdot\|)$, every interval is a connected set. Since $I_k$ is an interval, it is connected. Furthermore, a subset $S$ of a metric space $E$ is connected, if and only if there is no open cover $\{A,B\}$ of $S$ such that $S\cap A\neq\emptyset$, $S\cap B\neq\emptyset$ and $S\cap A\cap B=\emptyset$. Hence, by also considering that $I_k$ is connected, it is deduced that at most one set in $\{I_k\cap J_{\ell}:\ell\in L\}$ is non-empty.

So, exactly one set in $\{I_k\cap J_{\ell}:\ell\in L\}$ is non-empty, say $I_k\cap J_{\ell}$. Consequently, $I_k=I_k\cap J_{\ell}$, which in turns gives $I_k\subseteq J_{\ell}$.

In an analogous manner, starting from $J_{\ell}=\underset{k\in K}\bigcup(J_{\ell}\cap I_k)$, it is concluded that $J_{\ell}\subseteq I_k$. So, $I_k=J_{\ell}$, therefore the collections $\{I_k : k\in K\}$ and $\{J_{\ell} : \ell\in L\}$ coincide.

Proof of $\boldsymbol{\mathcal{B}(\mathbb{R})=\sigma(\pi(\mathbb{R}))}$

Let $\mathcal{O}$ be the family of open subsets in the Euclidean metric space $(\mathbb{R},\|\cdot\|)$. It will be shown that the Borel σ-algebra $\mathcal{B}(\mathbb{R})=\sigma(\mathcal{O})$ on $\mathbb{R}$ is equal to the σ-algebra $\sigma(\pi(\mathbb{R}))$ on $\mathbb{R}$ generated by $\pi (\mathbb{R})=\{(-\infty, x]:x\in\mathbb{R}\}$.

Firstly, it will be proved that $\pi (\mathbb{R})\subseteq\sigma(\mathcal{O})$. Consider a set $(-\infty, x]\in\pi (\mathbb{R})$. Note that $(-\infty, x]=\underset{n\in\mathbb{N}}{\bigcap}(-\infty,x+1/n)$. Indeed, for $y\in\underset{n\in\mathbb{N}}{\bigcap}(-\infty,x+1/n)$, it holds that $(\forall n\in\mathbb{N})y. Now assume that $x, in which case $y-x>0$, so by the Archimedean property, there exists $n_o \in\mathbb{N}$ such that $0<1/n_o , whence $x+1/n_o, which is a contradiction. Thus, $y\le x$, i.e. $y\in(-\infty,x]$. It has been established that $\underset{n\in\mathbb{N}}{\bigcap}(-\infty,x+1/n)\subseteq (-\infty,x]$. Obviously  $(-\infty,x]\subseteq\underset{n\in\mathbb{N}}{\bigcap}(-\infty,x+1/n)$ holds too.

Since $(-\infty,x+1/n)\in\mathcal{O}$ and $\sigma(\mathcal{O})$ is a σ-algebra on $\mathbb{R}$, the countable intersection $(-\infty, x]=\underset{n\in\mathbb{N}}{\bigcap}(-\infty,x+1/n)$ belongs to $\sigma(\mathcal{O})$.

With the help of lemma 1, $\pi (\mathbb{R})\subseteq\sigma(\mathcal{O})$ yields $\sigma(\pi (\mathbb{R}))\subseteq\sigma(\mathcal{O})=\mathcal{B}(\mathbb{R})$.

To prove the converse set inequality $\sigma(\mathcal{O}) \subseteq\sigma(\pi (\mathbb{R}))$, start by proving that $\mathcal{O}\subseteq\sigma(\pi (\mathbb{R}))$. Along these lines, consider an open set $G\in\mathcal{O}$. It is known from lemma 2 that $X$ can be written as a countable union of open intervals.

Let $(a, b)$ be one of these open intervals in the countable union. Recall from the proof of lemma 2 that $a, b$ can be infinite. Consequently, four cases are distinguished:

1. $a\in\mathbb{R}, b\in\mathbb{R}$. Set $(a, b)=\underset{n\in\mathbb{N}}{\bigcup}(a,b-\epsilon/n]$, where $\epsilon=(b-a)/2$. This case will be treated in detail.
2. $a=-\infty, b\in\mathbb{R}$. Set $(-\infty, b)=\underset{n\in\mathbb{N}}{\bigcup}(-\infty,b-\epsilon/n]$, where $\epsilon=(b-a)/2$.
3. $a\in\mathbb{R}, b=\infty$. Set $(a,\infty)=\underset{n\in\mathbb{N}}{\bigcup}(a,n]$.
4. $a=\infty, b=\infty$. Set $(-\infty, \infty)=(-\infty,c]\cup(c,\infty)$ for any real $c$, which leads back to case 3.

Case 1 is now treated in detail. The equality $(a, b)=\underset{n\in\mathbb{N}}{\bigcup}(a,b-\epsilon/n]$ is proved by observing that $y\in (a,b)\Leftrightarrow (\exists n_o \in\mathbb{N}) a. Leaving aside the remaining trivial deductions among these equivalences, it will be clarified that $y\in (a,b)\Rightarrow (\exists n_o \in\mathbb{N}) a. So, $y means $0 < (b-y)/\epsilon$. By the Archimedean property, there exists $n_o\in\mathbb{N}$ such that $0 < 1/n_o < (b-y)/\epsilon$, whence $y.

A set of the form $(a, u], a, can be written as $(a,u]=(-\infty, u]\cap(-\infty, a]^{c}$. So, the sets $(a,b-\epsilon/n], n\in\mathbb{N}$, appearing in $(a, b)=\underset{n\in\mathbb{N}}{\bigcup}(a,b-\epsilon/n]$ can be expressed as $(a,b-\epsilon/n]=(-\infty,b-\epsilon/n]\cap(-\infty, a]^{c}$. Thereby, $(-\infty,b-\epsilon/n]\in\pi(\mathbb{R})$ and $(-\infty,a]\in\pi(\mathbb{R})$ along with the fact that $\sigma (\pi(\mathbb{R}))$ is a σ-algebra lead to $(a,b-\epsilon/n]\in\sigma (\pi(\mathbb{R}))$. So, $(a,b)\in\sigma (\pi(\mathbb{R}))$, which implies $\mathcal{O}\subseteq\sigma(\pi (\mathbb{R}))$. Finally, $\sigma(\mathcal{O}) \subseteq\sigma(\pi (\mathbb{R}))$ by application of lemma 1, so $\mathcal{B}(\mathbb{R})=\sigma(\mathcal{O}) =\sigma(\pi (\mathbb{R}))$.

# Clarification on the σ-algebra generated by a set

This post elaborates on the concept of σ-algebra generated by a set, as presented in p. 17 of the book “Probability with martingales” by David Williams.

Let $\mathcal{C}$ be a class of subsets of a set $S$. It is stated in the book that the σ-algebra $\sigma(\mathcal{C})$ generated by $\mathcal{C}$ is the smallest σ-algebra on $S$ such that contains $\mathcal{C}$. Moreover, it is mentioned that $\sigma(\mathcal{C})$ is the intersection of all σ-algebras on $S$ that contain $\mathcal{C}$.

Three clarifications will be made. Let $\mathcal{G}:=\{\sigma\mbox{-algebra }\Sigma_i\mbox{ on }S\mbox{ with }\mathcal{C}\subseteq \Sigma_i:i\in I\}$ be the set of all σ-algebras on $S$ that contain $\mathcal{C}$.

Firstly, it will be shown that the intersection $\underset{i}{\bigcap}\Sigma_i$, which is the intersection of all σ-algebras on $S$ that contain $\mathcal{C}$, is itself a σ-algebra on $S$ that contains $\mathcal{C}$. Since $S\in\Sigma_i$ for all $i\in I$, it follows that $S\in\underset{i}{\bigcap}\Sigma_i$. Moreover, consider a set $F\in\underset{i}{\bigcap}\Sigma_i$. So $F\in\Sigma_i$ for all $i$, which means $F^{c}\in\Sigma_i$ for all $i$, so $F^{c}\in\underset{i}{\bigcap}\Sigma_i$. Let $\{F_j:j\in J\}$ be a countable collection of sets $F_j\in\underset{i}{\bigcap}\Sigma_i$. It follows that $F_j\in \Sigma_i$ for all $i\in I$ and all $j\in J$, so $\underset{j}{\bigcup}F_j\in\Sigma_i$ for all $i$, which gives $\underset{j}{\bigcup}F_j\in\underset{i}{\bigcap}\Sigma_i$. It has thus been confirmed that $\underset{i}{\bigcap}\Sigma_i$ is a σ-algebra on $S$. As $\mathcal{C}\subseteq\Sigma_i$ for all $i$, it becomes obvious that $\mathcal{C}\subseteq\underset{i}{\bigcap}\Sigma_i$, so the σ-algebra $\underset{i}{\bigcap}\Sigma_i$ on $S$ contains $\mathcal{C}$.

Consider now those σ-algebras $\Sigma_k\in\mathcal{G}$ with $\mathcal{C}\subseteq\Sigma_k$ that satisfy $\Sigma_k\subseteq\Sigma_i$ for all $\Sigma_i\in\mathcal{G}$ with $\mathcal{C}\subseteq\Sigma_i$. This is what is meant by “smallest σ-algebra $\Sigma$ on $S$ such that $\mathcal{C}\subseteq\Sigma$” in William’s book. Assume that there are two such distinct σ-algebras $\Sigma_k$ and $\Sigma_{\ell}$, i.e. $\Sigma_k\neq\Sigma_{\ell}$. Since $\Sigma_{\ell}\in\mathcal{G}$, it is then deduced that $\Sigma_k\subseteq\Sigma_{\ell}$. Similarly, $\Sigma_{\ell}\subseteq\Sigma_{k}$, so $\Sigma_k=\Sigma_{\ell}$, which is a contradiction. Hence, there is a single σ-algebra in $\mathcal{G}$ that contains $\mathcal{C}$ and is a subset of any other σ-algebra in $\mathcal{G}$ that also contains $\mathcal{C}$.

It has been shown that $\underset{i}{\bigcap}\Sigma_i\in\mathcal{G}$. Moreover, $\underset{i}{\bigcap}\Sigma_i$ satisfies $\underset{i}{\bigcap}\Sigma_i\subseteq\Sigma$ for any $\Sigma\in\mathcal{G}$. Uniqueness has also been established, so $\underset{i}{\bigcap}\Sigma_i$ is the only element in $\mathcal{G}$ that is a subset of all $\Sigma\in\mathcal{G}$. Put in words, $\underset{i}{\bigcap}\Sigma_i$ is the only σ-algebra on $S$ that contains $\mathcal{C}$ and is contained in any other σ-algebra $\Sigma$ on $S$ that satisfies $\mathcal{C}\subseteq\Sigma$.

$\underset{i}{\bigcap}\Sigma_i$ is called the σ-algebra generated by $\mathcal{C}$ and is denoted by $\sigma(\mathcal{C})$.

# Clarification on algebras and σ-algebras of a set

David Williams makes a terminological remark on algebras in p. 16 of his book “Probability with martingales”; he mentions that an algebra $\Sigma$ on a set $S$ (defined ordinarily as a collection of subsets of $S$ that contains $S$ and that is closed under set complementation and closed under finite set unions) is “a true algebra in the algebraists’ sense”.

This remark emphasises two aspects of the concept of algebra $\Sigma$ on a set $S$. Firstly, it means that $\Sigma$ is an algebra over a field $K$. The field $K$ contains exactly two elements, so it can be defined to be the subset $K=\{0,1\}$ of integers. According to the definition of algebra over a field, $\Sigma$ is a vector space equipped with a bilinear product. So any algebra (and consequently any σ-algebra) is a type of vector space of sets (subsets of $S$).

Secondly, consider the defining operations of an algebra $\Sigma$ on a set $S$. The vector space addition is defined to be the symmetric difference $A\Delta B:= (A\cup B)\setminus (A\cap B)$. The bilinear product, which turns the vector space to an algebra over $\{0,1\}$, is the set intersection $A\cap B$. It is straightforward to check that the product $A\cap B$ satisfies right and left distributivity and compatibility with scalars. The two operations (addition and bilinear product) defined from the product space $\Sigma\times\Sigma$ to the vector space $\Sigma$ are both symmetric.

In summary, any algebra (and any σ-algebra) is a vector space of sets equipped with two symmetric set operations.

# Proof: a and n are relatively prime iff ax≡b(modn) has a solution modulo n

The lemma in p. 231 of the book “A book of abstract algebra” by Charles C. Pinter states that if $a$ and $n$ are relatively prime, then the equation $ax\equiv b(\mbox{mod} n)$ has a solution modulo $n$. It is also stated below the lemma that if $a$ and $n$ are not relatively prime, then the equation $ax\equiv b(\mbox{mod} n)$ has no solution modulo $n$. By contraposition, the latter statement means that if the equation $ax\equiv b(\mbox{mod} n)$ has a solution modulo $n$, then $a$ and $n$ are relatively prime.

More generally, less emphasis is given on the converse of the lemma in the literature. The purpose of this post is to clarify that the equivalence holds. To start with, $a$ and $n$ are relatively prime if and only if their greatest common divisor satisfies $\mbox{gcd}(a, n)=1$. Along these lines, the lemma can then be stated in the following way; $\mbox{gcd}(a, n)=1$ if and only if the equation $ax\equiv b(\mbox{mod} n)$ has a solution modulo $n$.

Proofs of the fact that $\mbox{gcd}(a, n)=1$ means that the equation $ax\equiv b(\mbox{mod} n)$ has a solution modulo $n$ are available in Pinter’s book and elsewhere in the literature, so interest is in proving the converse. It will be shown that if the equation $ax\equiv b(\mbox{mod} n)$ has a solution modulo $n$, then $\mbox{gcd}(a, n)=1$.

If the congruence $ax\equiv b(\mbox{mod} n)$ has a solution, then it has a solution modulo $m$, where $m = n/\mbox{gcd}(a,n)$ (see for example theorem 2, p. 231, in Pinter’s book). Obviously, $m\le n$.

It will be shown that $m = n$. Towards this end, assume that $m < n$. So, it has been assumed that the congruence $ax\equiv b(\mbox{mod} n)$ has a solution modulo $n$. This means that it has a solution, therefore it has a solution modulo $m$.

Let $x\equiv c(\mbox{mod}m)$ be the solution modulo $m$ and $x\equiv r(\mbox{mod}n)$ the solution modulo $n$. Since $x = c$ is a solution of the congruence $ax\equiv b(\mbox{mod} n)$, it follows that $c\equiv r(\mbox{mod}n)$. The congruences $x\equiv r(\mbox{mod}n)$ and $c\equiv r(\mbox{mod}n)$ lead to $x\equiv c(\mbox{mod}n)$.

So, $x\equiv c(\mbox{mod}n)\Leftrightarrow x\equiv c(\mbox{mod}m)$. Notice now that $x\equiv c(\mbox{mod}n)$ implies $x-c\equiv 0(\mbox{mod}n)$, hence the coset equality $\overline{x-c}=\overline{0}_n$ holds in $\mathbb{Z}_n$, where $\overline{0}_n=\{\dots,-2n,-n,0,n,2n,\dots\}$. Similarly, $x\equiv c(\mbox{mod}m)$ yields $\overline{x-c}=\overline{0}_m$ in $\mathbb{Z}_m$, where $\overline{0}_m=\{\dots,-2m,-m,0,m,2m,\dots\}$. It has thus been concluded that $\overline{0}_n=\overline{0}_m$, which is a contradiction since for example $m\in\overline{0}_m$, while $m\not\in\overline{0}_n$ for $m. Thereby, $m=n$.

$m=n$ and $m = n/\mbox{gcd}(a,n)$ give $\mbox{gcd}(a, n)=1$, which completes the proof.

# Proof: characterisation of Jordan measurability

The following proof is a solution to exercise 1.1.5 of the book “An introduction to measure theory” by Terence Tao.

Some notation will be established before proceeding with the exercise. Let $\mathcal{E}(\mathbb{R}^d)$ be the set of elementary sets on $\mathbb{R}^d$. Denote by $\mathcal{L}(E):=\left\{m(A):A\in\mathcal{E}(\mathbb{R}^d),A\subseteq E\right\}$ and by $\mathcal{U}(E):=\left\{m(B):B\in\mathcal{E}(\mathbb{R}^d),E\subseteq B\right\}$ the sets of elementary measures of all elementary subsets and supersets of a bounded set $E\subseteq\mathbb{R}^d$, respectively. Let $m_{*}(E)=\sup{\mathcal{L}(E)}$ and $m^{*}(E)=\inf{\mathcal{U}(E)}$ be the Jordan inner and Jordan outer measures of $E$, respectively.

Exercise 1.1.5 requires to prove the equivalence of the following three statements, as a way of characterising Jordan measurability:

1. $E$ is Jordan measurable, which means that $m_{*}(E)=m^{*}(E)$.
2. For every $\epsilon>0$ there exist elementary sets $A\subseteq E\subseteq B$ such that $m(B\setminus A)\le\epsilon$.
3. For every $\epsilon>0$ there exists an elementary set $A$ such that $m^{*}(A\triangle E)\le\epsilon$.

It suffices to prove that $[1]\Rightarrow [2]\Rightarrow [3]\Rightarrow [1]$. To provide further practice and familiarity with Jordan inner and outer measures, it will be additionally shown how to prove $[1]\Rightarrow [3]\Rightarrow [2]\Rightarrow [1]$.

$\boxed{[1]\Rightarrow [2]}$

A reductio ad absurdum argument will be used. Assume that there exists $\epsilon_0>0$ such that for all elementary sets $A,B\in\mathcal{E}(\mathbb{R}^d)$ with $A\subseteq E\subseteq B$ the inequality $m(B\setminus A)>\epsilon_0$ holds.

Considering the set equality $B=A\cup (B\setminus A)$, it follows from $A\subseteq B$ that $m(B\setminus A)=m(B)-m(A)$. Hence, $m(B\setminus A)=m(B)-m(A)>\epsilon_0$.

So, $m(A)+\epsilon_0\le m^{*}(E)$, since $m(A)+\epsilon$ is a lower bound of $\mathcal{U}(E)$ and $m^{*}(E)=\inf{\mathcal{U}(E)}$. In turn, $m^{*}(E)-\epsilon_0$ is an upper bound of $\mathcal{L}(E)$ and $m_{*}(E)=\sup{\mathcal{L}(E)}$, therefore $m_{*}(E)\le m^{*}(E)-\epsilon_0$.

Thus, $m_{*}(E), which contradicts the assumption $m_{*}(E)=m^{*}(E)$.

$\boxed{[2]\Rightarrow [3]}$

Assume that there exists $\epsilon_0>0$ such that $\forall C\in\mathcal{E}(\mathbb{R}^d)$ holds $m^{*}(C\triangle E)>\epsilon_0$.

According to the assumed statement [2], for $\epsilon=\epsilon_0$, $\exists A, B\in\mathcal{E}(\mathbb{R}^d)$ with $A\subseteq E\subseteq B$ such that $m(B\setminus A)\le\epsilon_0$.

Pick $C=B$, so $m^{*}(B\triangle E)=m^{*}(C\triangle E)>\epsilon_0$. It follows from $E\subseteq B$ and $B\triangle E=(B\setminus E)\cup(E\setminus B)$ that $B\triangle E=B\setminus E$, so $m^{*}(B\setminus E)=m^{*}(B\triangle E)>\epsilon_0$.

$B\setminus E\subseteq B\setminus A$, since $A\subseteq E$. By also taking into account that $B\setminus A$ is elementary and the inequality $m(B\setminus A)\le\epsilon_0$, the conclusion is $m^{*}(B\setminus E)\le m(B\setminus A)\le\epsilon_0$.

A contradiction has been reached, as it has been deduced $m^{*}(B\setminus E)>\epsilon_0$ and $m^{*}(B\setminus E)\le\epsilon_0$ on the basis of the negation of statement [3].

$\boxed{[3]\Rightarrow [1]}$

Before proceeding with the proof of $[3]\Rightarrow [1]$, four lemmas will be proved.

Lemma 1

The Jordan inner measure $m_{*}(E)$ of any bounded set $E\subseteq\mathbb{R}^d$ is less than or equal to its Jordan outer measure $m^{*}(E)$, i.e. $m_{*}(E)\le m^{*}(E)$.

Proof of lemma 1

For any elementary sets $A,B$ with $A\subseteq E\subseteq B$, the set relation $A\subseteq B$ yields $m(A)\le m(B)$. This means that any $m(A)\in\mathcal{L}(E)$ is a lower bound of $\mathcal{U}(E)$, therefore $m(A)\le m^{*}(E)$. Since the last inequality holds for any $m(A)\in\mathcal{L}(E)$, it follows that $m^{*}(E)$ is an upper bound of $\mathcal{L}(E)$, thus $m_{*}(E)\le m^{*}(E)$.

Lemma 2

The elementary and Jordan measures of any elementary set $X$ coincide, that is $m_{*}(X)=m^{*}(X)=m(X)$.

Proof of lemma 2

It suffices to notice that $X\subseteq X$, whence $m^{*}(X)\le m(X)$ and $m(X)\le m_{*}(X)$, thereby $m^{*}(X)\le m_{*}(X)$. It is also known from lemma 1 that $m_{*}(X)\le m^{*}(X)$, so $m_{*}(X)=m^{*}(X)$. Finally, $m^{*}(X)\le m(X)$, $m(X)\le m_{*}(X)$ and $m_{*}(X)=m^{*}(X)$ yield $m(X)=m_{*}(X)=m^{*}(X)$.

Lemma 3

Let $E\subseteq\mathbb{R}^d$ be a bounded set and $B\in\mathcal{E}(\mathbb{R}^d)$ with $E\subseteq B$. It then holds that $m(B)-m_{*}(E)\le m^{*}(B\setminus E)$.

Proof of lemma 3

Recall that for a set $X\subseteq\mathbb{R}$ the following equivalences hold:

• $\ell=\inf{X}\Leftrightarrow(\forall\epsilon>0)(\exists x\in X)x<\ell+\epsilon$.
• $u=\sup{X}\Leftrightarrow(\forall\epsilon>0)(\exists x\in X)u-\epsilon < x$.

According to the former equivalence, for any $\epsilon>0$ there exists an elementary set $B\setminus E\subseteq C$ such that

$m(C).

By using the latter equivalence, there exists an elementary set $A\subseteq B$ such that

$m_{*}(B)-\epsilon /2.

It follows from lemma 2 and inequality (2) that $m(B)-m_{*}(E). $B\setminus E\subseteq C$ and $A\subseteq B$ lead to $A\setminus C\subseteq E$, so $m(A\setminus C)\le m_{*}(E)$, which in turn gives

$m(B)-m_{*}(E).

Inequality (3) connects the statement of lemma 3 to an analogous simpler statement for elementary sets; in particular, it will be shown that

$m(A)-m(C)\le m(A\setminus C)\ \ \ \ (4)$.

Indeed, $(A\setminus C)\cup (A\cap C)=A$, so $m(A)-m(A\cap C)=m(A\setminus C)$. Moreover, $A\cap C\subseteq C$, which means $m(A\cap C)\le m(C)$, therefore $m(A)-m(C)\le m(A)-m(A\cap C)$. Inequality (4) has thus been reached.

(3) and (4) yield $m(B)-m_{*}(E), which is combined with (1) to give

$m(B)-m_{*}(E).

For two reals $x, y$, if $(\forall\epsilon>0) x\le y+\epsilon$, then $x\le y$. This can be shown by assuming $x>y$, whence $0, contradiction. As inequality (5) holds for any $\epsilon>0$, the conclusion $m(B)-m_{*}(E)\le m^{*}(B\setminus E)$ follows.

Lemma 4

The Jordan outer measure of the union of two bounded sets $X\subseteq\mathbb{R}^d$ and $Y\subseteq\mathbb{R}^d$ is less than or equal to the sum of the Jordan outer measures of the two sets, i.e. $m^{*}(X\cup Y)\le m^{*}(X)+m^{*}(Y)$.

Proof of lemma 4

Let $V\subseteq\mathbb{R}^d,W\subseteq\mathbb{R}^d$ be elementary sets with $X\subseteq V,Y\subseteq W$.

Start by noticing the set equality $V\cup W=V\cup(W\setminus V)$. So, $m(V\cup W)=m(V)+m(W\setminus V)$. Moreover, $W\setminus V\subseteq W$ implies that $m(W\setminus V)\le m(W)$. Thus,

$m(V\cup W)\le m(V)+m(W)\ \ \ \ (6)$.

Inequality (6) holds for elementary sets, thereby it is a special case of lemma 4.

$m^{*}(X\cup Y)\le m(V\cup W)$, since $X\cup Y \subseteq V\cup W$. By taking into account inequality (6), $m^{*}(X\cup Y)\le m(V)+m(W)$. So, $m^{*}(X\cup Y)-m(V)$ is a lower bound of $\mathcal{U}(Y)$ and $m^{*}(X\cup Y)-m(V)\le m^{*}(Y)$. Furthermore, $m^{*}(X\cup Y)-m^{*}(Y)$ is a lower bound of $\mathcal{U}(X)$, so $m^{*}(X\cup Y)-m^{*}(Y)\le m^{*}(X)$, quod erat demonstrandum.

Proof of $\boldsymbol{[3]\Rightarrow [1]}$ using the lemmas

The main idea of the proof of $\boldsymbol{[3]\Rightarrow [1]}$ is to show that for any $\epsilon>0$ the inequality $m^{*}(E)-m_{*}(E)\le \epsilon$ holds.

It is known from statement [3] that for any $(\epsilon>0)(\exists X\in\mathcal{E}(\mathbb{R}^d))$ such that

$m^{*}(X\triangle E)\le\epsilon/3\ \ \ \ (7)$.

Using the relevant property of infimum, for any $\epsilon>0$ there exists an elementary superset $E\setminus X\subseteq Y$ of $E\setminus X$ such that

$m(Y).

Introduce the set $B:=X\cup Y$. Obviously $B$ is an elementary set and $E\subseteq B$, so $m^{*}(E)\le m(B)$. It thus follows from lemma 3 that

$m^{*}(E)-m_{*}(E)\le m^{*}(B\setminus E)\ \ \ \ (9)$.

$B\setminus E=(X\cup Y)\setminus E=(X\setminus E)\cup (Y\setminus E)$, so by lemma 4

$m^{*}(B\setminus E)\le m^{*}(X\setminus E)+m^{*}(Y\setminus E)\ \ \ \ (10)$.

$X\setminus E\subseteq X\triangle E\Rightarrow \mathcal{U}(X\triangle E)\subseteq\mathcal{U}(X\setminus E)$, so $m^{*}(X\setminus E)=\inf{\mathcal{U}(X\setminus E)}\le \inf{\mathcal{U}(X\triangle E)}=m^{*}(X\triangle E)$. Consequently, inequality (10) gives

$m^{*}(B\setminus E)\le m^{*}(X\triangle E)+m^{*}(Y\setminus E)\ \ \ \ (11)$.

In a similar way, $E\setminus X\subseteq X\triangle E$ implies

$m^{*}(E\setminus X)\le m^{*}(X\triangle E)\ \ \ \ (12)$.

Finally, $Y\setminus E\subseteq Y$ means that

$m^{*}(Y\setminus E)\le m(Y)\ \ \ \ (13)$.

All the components of the proof have been established. More concretely, inequalities (9), (11), (7), (13), (8) and (12) produce $m^{*}(E)-m_{*}(E)\le \epsilon$ for any $\epsilon>0$, so $m_{*}(E)=m^{*}(E)$.

$\boxed{[1]\Rightarrow [3]}$

Assume that $(\exists\epsilon_0>0)(\forall X\in\mathcal{E}(\mathbb{R}^d))m^{*}(X\triangle E)>\epsilon_0$.

As $m^{*}(E)=\inf{\mathcal{U}(E)}$, it is deduced that there exists an elementary set $E\subseteq B$ such that

$m(B).

Since $m_{*}(E)=\sup{\mathcal{L}(E)}$, there exists an elementary set $A\subseteq E$ such that

$m_{*}(E)-\epsilon_0/2.

$A\subseteq E\subseteq B\Rightarrow E\setminus A\subseteq B\setminus A$, so $m^{*}(E\setminus A)\le m(B\setminus A)$. Moreover, $A\subseteq B$ means $m(B\setminus A)=m(B)-m(A)$, hence

$m_{*}(E\setminus A)\le m(B)-m(A)\ \ \ \ (16)$.

Combining (14), (15) and (16) gives $m_{*}(E\setminus A). By the assumed statement [1], $m_{*}(E)=m^{*}(E)$, thus $m_{*}(E\setminus A)<\epsilon_0$. This is a contradiction, as the assumed negation of statement [3] gives $m_{*}(A\triangle E)=m_{*}(E\setminus A)>\epsilon_0$ for $A\subseteq E$.

$\boxed{[3]\Rightarrow [2]}$

The proof of $[3]\Rightarrow [2]$ is similar in spirit to the proof of $[3]\Rightarrow [1]$. It will be shown that for any $\epsilon >0$ there exist $A,B\in\mathcal{E}(\mathbb{R}^d)$ with $A\subseteq E\subseteq B$ such that $m(B\setminus A)\le \epsilon$.

It is known from statement [3] that for any $(\epsilon>0)(\exists X\in\mathcal{E}(\mathbb{R}^d))$ such that

$m^{*}(X\triangle E)\le\epsilon/4\ \ \ \ (17)$.

According to the relevant property of infimum, for any $\epsilon>0$ there exists an elementary set $Y$ with $E\setminus X\subseteq Y$ such that

$m(Y).

Consider the set $B:=X\cup Y$, which is an elementary set and $E\subseteq B$, so $m^{*}(E)\le m(B)$. By application of lemma 3,

$m(B)-m_{*}(E)\le m^{*}(B\setminus E)\ \ \ \ (19)$.

It is noted that $B\setminus E=(X\cup Y)\setminus E=(X\setminus E)\cup (Y\setminus E)$, so by lemma 4

$m^{*}(B\setminus E)\le m^{*}(X\setminus E)+m^{*}(Y\setminus E)\ \ \ \ (20)$.

$X\setminus E\subseteq X\triangle E$, so $m^{*}(X\setminus E)\le m^{*}(X\triangle E)$. Thus, inequality (20) leads to

$m^{*}(B\setminus E)\le m^{*}(X\triangle E)+m^{*}(Y\setminus E)\ \ \ \ (21)$.

Similarly, $E\setminus X\subseteq X\triangle E$ implies

$m^{*}(E\setminus X)\le m^{*}(X\triangle E)\ \ \ \ (22)$.

Moreover, $Y\setminus E\subseteq Y$ gives

$m^{*}(Y\setminus E)\le m(Y)\ \ \ \ (23)$.

Finally, there exists an elementary set $A$ with $A\subseteq E$ such that

$m_{*}(E)-\epsilon /4.

Inequalities (19), (21), (17), (23), (18) and (22) produce

$m(B)-m_{*}(E)\le 3\epsilon /4 \ \ \ \ (25)$.

Combining inequalities (24) and (25) confirms that for any $\epsilon > 0$ there exist elementary sets $A,B$ with $A\subseteq E\subseteq B$ such that $m(B)-m(A)\le \epsilon$, which completes the proof of $[3]\Rightarrow [2]$.

$\boxed{[2]\Rightarrow [1]}$

It is known from lemma 1 that $m_{*}(E)\le m^{*}(E)$. Assume the negation of statement [1], that is assume $m_{*}(E)\neq m^{*}(E)$. So, $m_{*}(E)< m^{*}(E)$.

Set $\epsilon_0=(m^{*}(E)-m_{*}(E))/2>0$. From the assumed statement [2], it is known that there exist $A,B\in\mathcal{E}(\mathbb{R}^d)$ with $A\subseteq E\subseteq B$ such that

$m(B\setminus A)=m(B)-m(A)\le \epsilon_0=(m^{*}(E)-m_{*}(E))/2 \ \ \ \ (26)$.

Note that $A\subseteq E\Rightarrow m(A)\le m_{*}(E)$ and $E\subseteq B\Rightarrow m^{*}(E)\le m(B)$, which leads to

$m^{*}(E)-m_{*}(E)\le m(B)-m(A) \ \ \ \ (27)$.

It follows from equation (26) that $2(m(B)-m(A))\le m^{*}(E)-m_{*}(E)$, which is combined with equation (27) to give $2(m(B)-m(A))\le m(B)-m(A)$, and finally $m(B)-m(A)\le 0$. Moreover, equation (27) yields $0 < 2\epsilon_0=m^{*}(E)-m_{*}(E)\le m(B)-m(A)$, i.e. $m(B)-m(A)>0$. Thus, a contradiction has been reached.

# Proof: uniqueness of elementary measure

The following proof is a solution to exercise 1.1.3 of the book “An introduction to measure theory” by Terence Tao.

A box $B\in\mathbb{R}^d$, $d\in\mathbb{N}$, is a Cartesian product $B:={\sf X}_{i=1}^d I_i$, where each interval $I_i$ is $I_i=(a, b)$ or $I_i=(a, b]$ or $I_i=[a, b)$ or $I_i=[a, b]$ for $a,b\in\mathbb{R}$ with $a\le b$. An elementary set $E=\cup_{i=1}^n B_i\subseteq\mathbb{R}^d$ is a finite union of disjoint boxes $B_i\in\mathbb{R}^d$. Let $\mathcal{E}(\mathbb{R}^d)$ denote the collection of elementary sets in $\mathbb{R}^d$. The measure $m:\mathcal{E}(\mathbb{R}^d)\rightarrow R^{+}\cup\left\{0\right\}$ is defined as $m(E)=\displaystyle\lim_{N\rightarrow\infty}\frac{1}{N^d}\#\left(E\cap\frac{1}{N}\mathbb{Z}^d\right)$, where $\#(\cdot)$ denotes set cardinality and $\displaystyle\frac{1}{N}\mathbb{Z}^d:=\left\{\frac{\mathbf{z}}{N}:\mathbf{z}\in\mathbb{Z}^d\right\}$.

Let $m^{'}:\mathcal{E}(\mathbb{R}^d)\rightarrow R^{+}\cup\left\{0\right\}$ be a function satisfying the non-negativity ($m^{'}(E)\ge 0$ for any elementary set $E$), finite additivity ($m^{'}(\displaystyle\cup_{i=1}^n E_i)\le \sum_{i=1}^{n}m^{'}(E_i)$ for disjoint elementary sets $E_i$) and translation invariance ($m^{'}(E+\mathbf{x})=m^{'}(E)$ for any elementary set $E$ and any $\mathbf{x} \in \mathbb{R}^d$) properties.

It will be shown that there exists a positive constant $c\in\mathbb{R}^+$ such that $m^{'}=cm$, i.e. the functions $m^{'}$ and $m$ are equal up to a positive normalization constant $c$.

Observe that $\left[0,1\right)=\displaystyle\cup_{i=0}^{n-1}\left[\frac{i}{n},\frac{i+1}{n}\right)$. Due to translation invariance, $m^{'}\left(\displaystyle\left[\frac{i}{n},\frac{i+1}{n}\right)\right)=m^{'}\left(\displaystyle\left[\frac{i}{n},\frac{i+1}{n}\right)-\frac{i}{n}\right)=m^{'}\left(\left[0,\frac{1}{n}\right)\right)$. Using finite additivity, it follows that $m^{'}\left(\left[0,1\right)^d\right)=n^dm^{'}\left(\left[0,\frac{1}{n}\right)^d\right)$. So, $m^{'}\left(\left[0,\frac{1}{n}\right)^d\right)=\displaystyle\frac{c}{n^d}$ for $c:=m^{'}\left(\left[0,1\right)^d\right)$. Since $m\left(\left[0,\frac{1}{n}\right)^d\right)=\displaystyle\frac{1}{n^d}$, it follows that $m^{'}\left(\left[0,\frac{1}{n}\right)^d\right)=c m\left(\left[0,\frac{1}{n}\right)^d\right)$.

This result generalizes to intervals $\left[0,q\right)$ for any rational $q=\displaystyle\frac{s}{n}$ with $s\in\mathbb{N}$ and$n\in\mathbb{N}$. Since $\left[0,q\right)=\displaystyle\cup_{i=0}^{s-1}\left[\frac{i}{n},\frac{i+1}{n}\right)$, finite additivity and translation invariance lead to $m^{'}\left(\left[0,q\right)^d\right)=c m\left(\left[0,q\right)^d\right)=cq^d$.

It will be shown that the result holds also for intervals $\left[0,p\right)$ for any irrational $p\in\mathbb{P}$.

The set of rationals is dense, which means that $(\forall \epsilon>0)(\forall x\in\mathbb{R})(\exists q\in\mathbb{Q})|x-q|<\epsilon$. For some irrational $x=p$ and for each $n\in\mathbb{N}$, set $\epsilon=\displaystyle\frac{1}{n}$, so $(\forall n\in\mathbb{N})(\exists q_n\in\mathbb{Q})|p-q_n|<\displaystyle\frac{1}{n}$. Pick some $n_0\in\mathbb{N}$ with $n_0>\displaystyle\frac{2}{p}$. For all $n\in\mathbb{N}$ with $n>n_0$, it holds that $\displaystyle \frac{2}{n}<\frac{2}{n_0}, whence $0<\displaystyle\frac{1}{n}. So, $(\exists n_0\in\mathbb{N})(\forall n\in\mathbb{N})$ with $n>n_0$, it is true that $0 and consequently $\left[0,q_n-\displaystyle\frac{1}{n}\right)^d\subseteq\left[0,p\right)^d\subseteq\left[0,q_n+\displaystyle\frac{1}{n}\right)^d$.

For any two elementary sets $E\subseteq F$, it can be shown that $m^{'}(E)\le m^{'}(F)$ via the equality $F=E\cup(F\setminus E)$, non-negativity and finite additivity. Hence, $m^{'}\left(\left[0,q_n-\displaystyle\frac{1}{n}\right)^d\right)\le m^{'}\left(\left[0,p\right)^d\right)\le m^{'}\left(\left[0,q_n+\displaystyle\frac{1}{n}\right)^d\right)$.

Since $q_n\pm\displaystyle\frac{1}{n}$ are rationals, it is deduced that $m^{'}\left(\left[0,q_n\pm\displaystyle\frac{1}{n}\right)^d\right)=c\left(q_n\pm\displaystyle\frac{1}{n}\right)^d$. Thus, $c\left(q_n-\displaystyle\frac{1}{n}\right)^d\le m^{'}\left(\left[0,p\right)^d\right)\le c\left(q_n+\displaystyle\frac{1}{n}\right)^d$.

$0 gives $0, hence the sandwich theorem yields $\displaystyle\lim_{n\rightarrow\infty}q_n=p$.

Combining $c\left(q_n-\displaystyle\frac{1}{n}\right)^d\le m^{'}\left(\left[0,p\right)^d\right)\le c\left(q_n+\displaystyle\frac{1}{n}\right)^d$ and $\displaystyle\lim_{n\rightarrow\infty}q_n=p$ gives $cp^d\le m^{'}\left(\left[0,p\right)^d\right)\le cp^d$, so $m^{'}\left(\left[0,p\right)^d\right)=cp^d$ for any irrational $p$.

This effectively completes the proof. There remains to show that $m^{'}=cm$ is true for Cartesian products of unequal intervals $\left[0,x_i\right)$ in each coordinate $i=1,2,\dots,d$, for any $x_i\in\mathbb{R}$, and for any subinterval of the real line. These are all trivial given the existing foundations.