# 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 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.