Exercise 1.3.12 (Upper Lebesgue integral and outer Lebesgue measure)

We have to demonstrate that

Any simple integral from the left-hand side set will always have a form

for some measurable disjoint sets $I_1,\dots ,I_k\subseteq {ℝ}^d$ and reals $c_1,\dots ,c_k\in ℝ,c_i\ne 0$. Notice that due to the nature of the indicator function we can deduce two facts, one about $I_i$ and another about $c_i$. First, if $h$ majorizes $1_E$, then $I_1\cup \dots \cup I_K$ must contain $E$ (otherwise $h(x)=0\le 1=1_E(x)$ for $x\in E∕\bigcup <!--\; FUNCTION\; APPLICATION\; -->I_i$). Second, if the simple integral of $h=c_1{1}_{I_1}+\dots +c_k{1}_{I_k}$ is sufficiently close to $\underline{{\int }_{{ℝ}^d}}{1}_E$, then the simple integral of $h^{\prime }=1_{I_1}+\dots +1_{I_k}$ is even closer to $\underline{{\int }_{{ℝ}^d}}{1}_E$ by $h^{\prime }\ge 1_E$ and $h\ge h^{\prime }$.

With this in mind, by the countable axiom of choice we pick a sequence ${(h_n)}_{n\in \mathbb{N}}$ of simple functions majorizing $1_E$ such that

By the argument in the previous paragraph we safely assume that the coefficients of $h_n=1_{I_1^{(n)}}+\dots +1_{I_{k_n}^{(n)}}$ are always 1’s. Rewriting the above using the definition of the simple integral, we have

For each $n\in \mathbb{N}$ denote $E_n:=I_1^{(n)}\cup \dots \cup I_{k_n}^{(n)}$. Then $E_n$ is measurable, and we have

${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{E}_i$ is a Lebesgue measurable set; we now have to demonstrate that

by Exercise 1.2.7 we can find an open set $U$ containing ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{E}_i$ such that $m(U)-m\left({\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^{\infty }{E}_i\right)\le 𝜖$. By Lemma 1.2.11 we can express $U$ as a countable union of disjoint boxes ${(B_n)}_{n\in \mathbb{N}}$; by Lemma 1.2.9 the measure of $U$ is

But this means that we have found an element from the right-hand side such that

Taking $𝜖\to \infty$ in the inequality below we obtain the desired result.

We have to demonstrate

” $\le$”

Any simple function from the right-hand side will always have the form

with the $F_i$ disjoint by MTE 3.2b for some measurable disjoint sets ${\{F_i\}}_{i=1}^k$ and positive reals ${\{c_i\}}_{i=1}^k$.

Notice that if $h$ majorizes $1_E$, then $E\subseteq {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^k{F}_i$, because if $(E\setminus {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^k{F}_i)$ is non-empty, then for $x\in (E\setminus {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^k{F}_i)$, we have $1_E(x)=1>0=h(x)$.

Second, if in the representation of $h$ there are any $F_i$ with $F_i\cap E=\varnothing$, we can just ignore those $F_i$ and still have a simple function majorizing $1_E$.

Third, note that if the simple integral of $h={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^k{c}_i{1}_{F_i}$ is sufficiently close to ${\bar{\int }}_{{ℝ}^d}{1}_E\mathit{dx}$, then each $c_i\ge 1$. Now the simple integral of $h^{\prime }={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^k{1}_{F_i}$ is even closer to ${\int }_{{ℝ}^d}$ since $h^{\prime }\ge 1_E$ and $h\ge h^{\prime }$. So WLOG, let us assume $F_i\cap E\ne \varnothing$ and all $c_i$ equal 1 and the $F_i$ are disjoint.

Let $𝜖>0$ be arbitrary. Choose an arbitrary simple function $h$ majorizing $1_E$ with the properties above. Then:

Since the sets $F_j$ are measurable, $m(F_j)=m^{\text{∗}}(F_j)$, and there are $k$ sequences of boxes ${\{B_i^{(j)}\}}_{i=1}^{\infty }$ for $1\le j\le k$ covering each $F_j$ with

This gives us

Since the union of countable index sets is countable again, we can subsume ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^k{\bigcup }_{i=1}^{\infty }{B}_i^{(j)}$ under one index set. That is, every simple function $h$ majorizing $1_E$ defines a box cover ${\{B_n\}}_{n=1}^{\infty }$ covering $E$ such that

Then:

Taking $𝜖\to 0$ gives us

” $\ge$”

For the other direction, assume a box cover ${\{B_n\}}_{n=1}^{\infty }$ of $E$, that is, $E\subseteq {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n$. Then $1_{{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{B}_n}$ is a simple function majorizing $1_E$, and:

Taking infimum on both sides gives us:

Thus, we are done.