Exercise 1.2.13 (Dominated convergence theorem for measurable sets)

Our goal is to demonstrate that $E$ can be represented as a combination of countable unions/intersections of $E_n$’s. We notice two facts:

(i)

(Constant tail of the sequence) First notice that due to the binary nature of ${(1_{E_n}(x))}_{n\in \mathbb{N}}$ the pointwise convergence $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{1}_{E_n}(x)=1_E(x)$ is equivalent to the condition that the sequence stabilises $1_{E_n}(x)=1_E(x)$ after some $n\ge N\in \mathbb{N}$.
Pick an arbitrary $x\in E$. Notice that by the binary nature of $1_{E_n}(x)\in \{0,1\}$ the pointwise convergence condition $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{1}_{E_n}(x)=1_E(x)=1$ implies that we only have finite number of $E_n$’s such that $1_{E_n}(x)=0$; otherwise we could build a limit inferior ${\mathrm{liminf}<!--\; FUNCTION\; APPLICATION\; -->}_{n\to \infty }{1}_{E_n}=0$ - a contradiction. In other words, we can find a $N\in \mathbb{N}$ such that after this element $\forall <!--\; FUNCTION\; APPLICATION\; -->n\ge N$ the sequence stabilises $1_{E_n}=1$. Conversely, for $x\in {ℝ}^d$ if there is an $N\in \mathbb{N}$ after which the sequence eventually becomes a constant $1$-sequence, then we must have $1_E(x)=1$ by the limit assumption.
Thus, using the first criterion we can represent $E$ as $$\begin{array}{llll}\hfill E& =\{x\in {ℝ}^d:x\in E\}& \hfill & \\ \hfill & =\bigcup _{N\in \mathbb{N}}\left\{x\in {ℝ}^d:1_{E_n}(x)=1\text{ for all }n\ge N\right\}& \hfill & \\ \hfill & =\bigcup _{N\in \mathbb{N}}\bigcap _{n\ge N}\left\{x\in {ℝ}^d:1_{E_n}(x)=1\right\}& \hfill & \\ \hfill & =\bigcup _{N\in \mathbb{N}}\bigcap _{n\ge N}{E}_n& \hfill & \end{array}$$

(ii)

(Limit point) Another equivalent formulation is that $1_E(x)$ is the only limit point of ${(1_{E_n}(x))}_{n\in \mathbb{N}}$, i.e., we have the unique condition $\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}$ $\exists <!--\; FUNCTION\; APPLICATION\; -->N\ge n$ such that $1_{E_n}(x)=1_E(x)$.
Thus, using this second creterion we write $$\begin{array}{llll}\hfill E& =\{x\in {ℝ}^d:x\in E\}& \hfill & \\ \hfill & =\bigcap _{n\in \mathbb{N}}\left\{x\in {ℝ}^d:1_{E_N}(x)=1\text{ for some }N\ge n\right\}& \hfill & \\ \hfill & =\bigcap _{n\in \mathbb{N}}\bigcup _{N\ge n}\left\{x\in {ℝ}^d:1_{E_N}(x)=1\right\}& \hfill & \\ \hfill & =\bigcap _{n\in \mathbb{N}}\bigcup _{N\ge n}{E}_N& \hfill & \end{array}$$

Thus, (in any of the above two examples) we have managed to represent $E$ as a countable union of countable intersections of measurable sets; by Lemma 1.1.13 $E$ is measurable as well.

These both representation help us derive the Dominated Convergence Theorem from the Monotone Convergence Theorem. In the first case notice that due to ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{n\ge 1}{E}_n\subseteq {\bigcap }_{n\ge 2}{E}_n\subseteq \dots$ we can use the upward convergence

In the second case, due to ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n\ge 1}{E}_n\supseteq {\bigcup }_{n\ge 2}{E}_n\supseteq \dots$ we can use the downward convergence

Cobining both facts we obtain

Note that the inverse counterexample from Exercise 1.2.11 (removing blocks from $[0,+\infty )$ countably by $n$ steps) is also applicable here.