Exercise 1.4.24 (Dominated convergence theorem for measurable sets)

Question

Let $(X,\mathcal{B},\mu)$ be a measure space. Let $\left(Eight)_{n\in\mathbb{N}}$ be a sequence of $\mathcal{B}$-measurable sets that converge to another set $E$ (i.e., $1_{E_n}	o 1_E$).
\begin{enumerate}[label=(oman*)]
\item Show that $E$ is also $\mathcal{B}$-measurable
\item If there exists a $\mathcal{B}$-measurable set $F\subseteq X$ of finite measure which contains all of the $\left(Eight)_{n\in\mathbb{N}}$, then $\lim_{n	o\infty}\mu(E_n)=\mu(E)$.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(\roman*)]
\item To show that $E$ is measurable, we argue that due to the pointwise convergence $E$ can be represented as a limit inferior (or limit superior) of $\left(E\right)_{n\in\mathbb{N}}$, i.e., as a countable union/intersection of the measurable sets:
      $$E = \bigcup_{n=1}^{\infty} \bigcap_{N=n}^{\infty} E_n = \bigcap_{n=1}^{\infty} \bigcup_{N=n}^{\infty} E_n$$
      We shortly verify the first equality. Pick an arbitrary $x\in E$. Then there exists a $n\in\mathbb{N}$ such that for all $N \geq n$ we have $1_{E_n}=1_E$ (just set $\epsilon < 1$). In other words, for all $N\geq n: x \in E_n$. Thus, $x\in \bigcap_{N=n}^{\infty}E_n$ and is contained in the left-hand side. Now pick an arbitrary $x \in \bigcup_{n=1}^{\infty} \bigcap_{N=n}^{\infty} E_n$, i.e., for some $n\in\mathbb{N}$ we have $x\in \bigcap_{N=n}^{\infty} E_n$. If we had $x\notin E$, then there would exist $N\in\mathbb{N}$ such that $1_{E_N}(x)=0$ - a contradiction. The second equality follows in the similar manner.
\item Notice that by measurability of $E,E_n$ we can use finite additivity to see that both $\mu(E)= \mu(E/E_n) + \mu(E \cap E_n)$ and $\mu(E_n) = \mu(E/E_n) + \mu(E_n \cap E)$ holds. Thus,
      \begin{align*}
        \lim_{N\to\infty}\mu(E)-\mu(E_N)&= \lim_{N\to\infty} \mu(E/E_n) - \mu(E_n/E)\
                                        &\leq \lim_{N\to\infty} \mu(E/E_n) + \mu(E_n/E) \&= \lim_{N\to\infty} \mu(E \triangle E_n)\
                                        &\leq \lim_{N\to\infty} \mu\left( \bigcup_{n=N}^{\infty}E_n \triangle E\right)
      \end{align*}
      Since $\bigcup_{n=N}^{\infty}E_n \triangle E \supseteq \bigcup_{n=N+1}^{\infty}E_n \triangle E$, by the downwards version of \href{./exercise_1-4-23}{Exercise 1.4.23}, we have
      \begin{align*}
       \lim_{N\to\infty}\mu(E)-\mu(E_N) &= \mu\left( \bigcap_{N=1}^{\infty}\bigcup_{n=N}^{\infty}E_n \triangle E\right) 
      \end{align*}
      We now argue that the last term is an empty set.
      \begin{align*}
        \bigcap_{N=1}^{\infty}\bigcup_{n=N}^{\infty}E_n \triangle E &= \bigcap_{N=1}^{\infty}\left[\left(\bigcup_{n=N}^{\infty}E_n/E\right) \cup \left(\bigcup_{n=N}^{\infty} E/E_n\right)\right] = \bigcap_{N=1}^{\infty}\left[\left(\bigcup_{n=N}^{\infty}E_n/E\right) \cup \left(E / \bigcap_{n=N}^{\infty} E_n\right)\right]\[5pt]
                                                                    &= \left[\bigcap_{N=1}^{\infty}\bigcup_{n=N}^{\infty}E_n/E\right] \cup \left[\bigcap_{N=1}^{\infty}\left(E / \bigcap_{n=N}^{\infty} E_n\right)\right] = \left[\bigcap_{N=1}^{\infty}\bigcup_{n=N}^{\infty}E_n/E\right] \cup \left[E / \bigcup_{N=1}^{\infty}\bigcap_{n=N}^{\infty} E_n\right]\[5pt]
                                                                    &= [E/E] \cup [E/E] = \emptyset
      \end{align*}
      From this, the assertion follows.
\end{enumerate}

Exercise 1.4.24 (Dominated convergence theorem for measurable sets)

Answers

Comments

Exercise 1.4.24 (Dominated convergence theorem for measurable sets)

Answers

Comments

Add answer