Exercise 1.2.11 (Monotone convergence theorem for measurable sets)

Question

\begin{enumerate}[label=(oman*)]
	\item (Upward monotone convergence) Let $E_1 \subseteq E_2 \ldots \subseteq \mathbb{R}^d$ be a countable non-decreasing sequence of Lebesgue measurable sets. Show that
		$$m\left(\bigcup_{n=1}^{\infty}E_night) = \lim_{n	o\infty} m(E_n)$$
	\item (Downward monotone convergence) Let $\mathbb{R}^d \supseteq E_1 \supseteq E_2 \ldots$ be a countable non-increasing sequence of Lebesgue measurable sets. Suppose that at least one of the $m(E_n)$ is finite. Show that
		$$m\left(\bigcap_{n=1}^{\infty}E_night) = \lim_{n	o\infty} m(E_n)$$
\end{enumerate}
Give a counterexample to the second part to show that the assumption that at least one $m(E_n)$ is finite cannot be dropped.

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(oman*)]
	\item Notice that we have the identity $\bigcup_{n=1}^{\infty}E_n = \bigcup_{n=1}^{\infty}E_n / \left(\bigcup_{i=1}^{n-1} E_i ight)$. Pick an arbitrary $x$ from the right-hand side; then, $x \in E_n / \bigcup_{i=1}^{n-1}E_i \subseteq E_n$ for some $n\in \mathbb{N}$. Similarly, $x$ from the left-hand side implies that there exists a finite $n := \min\{i\in\mathbb{N}: x \in E_i\}$. We then have $x\in E_n / E_{n-1} = E_n / \left(\bigcup_{i=1}^{n-1} E_i ight)$.
ewline      Applying this identity we get, by the countable additivity of the Lebesgue measure,
		\begin{align*}
m\left(\bigcup_{n=1}^{\infty}E_night) &= m\left(\bigcup_{n=1}^{\infty}E_n/\bigcup_{i=1}^{n-1}E_iight) = m\left(\bigcup_{n=1}^{\infty}E_n/E_{n-1}ight)\ &= \sum_{n=1}^{\infty} m\left(E_{n}/E_{n-1}ight)  = \sum_{n=1}^{\infty} m\left(E_{n}ight) - m\left(E_{n-1}ight).
\end{align*}
		By the law of telescoping series (cf. Lemma 7.2.15 from \href{../analysis_I}{Analysis I}), this series converges to $\lim_{n	o\infty} m(E_n)$.
	\item We combine two facts:
	\begin{itemize}
		\item On one hand, since $\bigcap_{n=1}^{\infty}E_n \subseteq E_1$ we have        
			$$m\left(E_1/\bigcap_{n=1}^{\infty}E_night) = m\left(E_1ight) - m\left(\bigcap_{n=1}^{\infty}E_night)$$
		\item On the other hand, note that $E_1/\bigcap_{n=1}^{\infty}E_n = \bigcup_{n=1}^{\infty} E_n/E_{n+1}$. We prove this. Pick $x \in E_1/\bigcap_{n=1}^{\infty}E_n$. Since $x$ is not contained in every $E_n$, then by the nestedness property there should be the first $n_0$ such that $x\in E_{n_0+1}$ and $x 
otin E_{n_0}$. Hence, $x\in \bigcup_{n=1}^{\infty} E_n/E_{n+1}$. Conversely, let $x \in \bigcup_{n=1}^{\infty} E_n/E_{n+1}$. Then, $x \in E_n/E_{n+1}$ for some $n\in\mathbb{N}$. Firstly, since $E_n \subseteq E_1$ we have $x\in E_1$. Second, since $x
otin E_{n+1}$ we have $x 
otin \bigcap_{n=1}^{\infty}E_n$. Thus, $x \in E_1 / \bigcap_{n=1}^{\infty}E_n $ as desired. Using this fact, we see        
\begin{align*}
m\left(E_1/\bigcap_{n=1}^{\infty}E_night) &= \sum_{n=1}^{\infty} m\left(E_night) - m\left(E_{n+1}ight)\
&= \lim_{N	o\infty} \left(\sum_{n=1}^{N} m\left(E_night) - m\left(E_{n+1}ight) ight)\
& = m(E_1) - \lim_{N	o\infty} m(E_{N+1})
\end{align*}
where in the last step we have used the telescoping series property again.
		\item Combining both equations we get
			$$m\left(\bigcap_{n=1}^{\infty}E_night) = \lim_{n	o\infty} m(E_{n}).$$
		\end{itemize}
	\item Consider the sequence $E_n := \mathbb{R}_+/[0,n]$. We have $m\left(\bigcap E_night) = m(\emptyset) = 0$. On the other hand, $\forall n\in\mathbb{N}: m(E_n) = \infty$; thus, the sequence of measures does not converge.
\end{enumerate}

Exercise 1.2.11 (Monotone convergence theorem for measurable sets)

Answers

Comments

Exercise 1.2.11 (Monotone convergence theorem for measurable sets)

Answers

Comments

Add answer