# 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.

# 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.

# 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.