Exercise 0.24 (Extended-real measurable functions)

Question

Let $\left(\Omega,\mathcal{F}ight)$ be a measurable space and $\left(\mathbb{R}^*,\mathcal{B}ight)$ be the extended reals with the Borel $\sigma$-algebra. Let $f,g:\Omega 	o [-\infty,+\infty]$ be functions taking values in the extended reals.
\begin{enumerate}
	\item Show that $f$ is measurable if and only if the sets $\left\{\omega \in \Omega: f(\omega) \leq tight\}$ are measurable for all reals $t$.
	\item Show that $f=g$ if and only if $\{\omega \in \Omega: f(\omega) \leq t\} = \{\omega \in \Omega: g(\omega) \leq t\}$ for all reals $t$.
	\item If $f_1,f_2,\ldots : \Omega 	o [-\infty,+\infty]$ are measurable, show that $\sup_{n\in\mathbb{N}}f_n$, $\inf_{n\in\mathbb{N}}f_n$, $\limsup_{n\in\mathbb{N}}f_n$ and $\liminf_{n\in\mathbb{N}}f_n$ are all measurable.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}
	\item[$\implies$] Suppose that $f$ is measurable. Since any closed set is measurable, then the inverse image $f^{-1}([-\infty,t])$ of the closed set $[-\infty,t]$ under the measurable function $f$ must be measurable in $(\Omega,\mathcal{B})$.
ewline
		[$\impliedby$] We use the measurable induction (Remark 1.4.15 from \href{../an_introduction_to_measure_theory}{Measure Theory}) with the property $P(S) = f^{-1}(S) \in \mathcal{F}$. We use the open sets of $\mathbb{R}^*$ as a generating set for $\mathcal{B}$. Let $O \in \mathcal{O}(\mathbb{R}^*)$. \href{https://math.stackexchange.com/questions/2692405/every-open-set-in-the-extended-real-line-overline-mathbb-r-is-a-countable}{Recall} that all open subsets of the extended real line can be written as countable unions of open intervals. Any open interval of the extended real line has either one of the following forms:
		\begin{itemize}
			\item $[-\infty,a)$ for $a \in \mathbb{R}$
ewline
				The inverse image under $f$ is measurable since$$f^{-1}\left([-\infty,a)ight) = f^{-1}\left(\bigcup_{n=1}^{\infty}\left[-\infty,a-\frac{1}{n}ight]ight) = \bigcup_{n=1}^{\infty} f^{-1}\left(\left[-\infty,a-\frac{1}{n}ight]ight)$$ and the countable union of measurable sets (by assumption) is measurable.
			\item $(b,+\infty]$ for $b \in \mathbb{R}$.
ewline
				The inverse image under $f$ is measurable since $f^{-1}\left((b,+\infty]ight) = f^{-1}\left(\mathbb{R}^*/[-\infty,b]ight) = f^{-1}\left(\mathbb{R}^*ight) \setminus f^{-1}\left([-\infty,b]ight)$. The last term is measurable since by the theorem assumption, it is a set difference of two measurable sets.
			\item $(a,b)$ for $a,b \in \mathbb{R}^*$
ewline
				This follows by representing $(a,b)$ as $(a,+\infty)\cap [-\infty,b)$.
		\end{itemize}
		Since any open set is a countable union of the above, it's inverse image must be measurable as well. In other words, for all $$O\in\mathcal{O}(\mathbb{R}^*): f^{-1}(O) \in \mathcal{F}.$$ Finally, we demonstrate that the above is a $\sigma$-algebra property:
		\begin{enumerate}
			\item $P(\emptyset) = f^{-1}(\emptyset) \in \mathcal{F}$ is true, since $f^{-1}(\emptyset) = \emptyset$ and the empty set is always contained in any $\sigma$-algebra $\mathcal{F}$.
			\item Suppose that $f^{-1}(S) \in \mathcal{F}$ is true for some $S \subseteq Y$. Using the set-theoretic property from the \href{../analysis_I/exercise_3-4-4}{Exercise 3.4.4} of \href{../analysis_I}{Analysis I} we see that $f^{-1}(\mathbb{R}^*/S) = f^{-1}(\mathbb{R}^*) / f^{-1}(S) = \Omega \setminus f^{-1}(S).$ Since $\mathcal{F}$ is a $\sigma$-algebra that contains $f^{-1}(S)$, it must also contain its complement.
			\item Let $S_1,S_2,\ldots \subseteq X$ be such that $P(S_i) = f^{-1}(S_i) \in \mathcal{F}$ is true for all $i \in \mathbb{N}$. Then, we also have by \href{../analysis_I/exercise_3-4-3}{Exercise 3.4.3} of \href{../analysis_I}{Analysis I} that $f^{-1}\left(\bigcup_{i=1}^{\infty}S_iight) = \bigcup_{i=1}^{\infty}f^{-1}\left(S_iight) \in \mathcal{F}$ is true as well since $\mathcal{F}$ contains the countable unions of any of its elements.
		\end{enumerate}
		Hence, we have $\forall S \in \mathcal{B}=\langle\mathcal{O}(\mathbb{R}^*)angle: f^{-1}(S) \in \mathcal{F}$.
	\item[$\implies$] This is obvious.
ewline
		[$\impliedby$] Consider an arbitrary $\overline{\omega}$ from the set $\left\{\omega \in \Omega: f(\omega) 
eq g(\omega)ight\}$. For this $\overline{\omega}$ we have either $f(\overline{\omega}) < g(\overline{\omega})$ or $f(\overline{\omega}) > g(\overline{\omega})$. Without the loss of generality, consider the first case. Define 
		$$f(\overline{\omega}) < t:=\frac{f(\overline{\omega})+g(\overline{\omega})}{2} < g(\overline{\omega}).$$
		We then see that
		$$\overline{\omega} \in \left\{\omega \in \Omega: f(\omega) < t ight\} \qquad\overline{\omega} 
ot\in \left\{\omega \in \Omega: g(\omega) < t ight\}$$
		which is a contradiction to the theorem assumption. In other words, $\left\{\omega \in \Omega: f(\omega) 
eq g(\omega)ight\} = \emptyset$.
	\item We now consider the limiting functions.
		\begin{itemize}
			\item $f(x) \mapsto \sup\left\{f_n(x) : n\in\mathbb{N}ight\}$
ewline
				Let $E\in\mathcal{B}$ be a measurable set. Consider $f^{-1}([-\infty,t])$ for some arbitrary $t\in\mathbb{R}$. Then it is easy to verify using logical operators that $$f^{-1}\left([-\infty,t]ight) = \bigcap_{n=1}^{\infty}f_n^{-1}\left([-\infty,t]ight).$$ $[\subseteq]$ Let $\omega \in f^{-1}\left([-\infty,t]ight)$. Then $\forall n \in \mathbb{N}: f_n(\omega) \leq f(\omega) \leq t$ by definition, and so $\omega \in f_n^{-1}\left([-\infty,t]ight)$ for all $n\in\mathbb{N}$.
ewline
				$[\supseteq]$ Let $\omega \in \bigcap f_n^{-1}\left([-\infty,t]ight)$. Then $\forall n\in\mathbb{N}: f_n(\omega)\leq t$ and so $\sup_{n\in\mathbb{N}}f_n(\omega) \leq t$ as well.
ewline
				Since a countable union of measurable function is again measurable, the set $f^{-1}\left([-\infty,t]ight)$ must be measurable as well.
			\item $f(x) \mapsto \inf\left\{f_n(x) : n\in\mathbb{N}ight\}$
ewlineFollows similarly by observing that $$f^{-1}\left([-\infty,t]ight) = \bigcup_{n=1}^{\infty}f_n^{-1}\left([-\infty,t]ight).$$
			\item $f(x) \mapsto \limsup\left\{f_n(x) : n\in\mathbb{N}ight\}$
ewlineBy Definition 6.4.6 from \href{../analysis_I}{Analysis I} we can write $\limsup$ as $$\inf_{N=m}^{\infty} \sup_{n=N}^{\infty} f_n(x).$$ In other words, $\limsup$ is a combination of $\inf$ and $\sup$ and so must be measurable by the previous parts.
			\item $f(x) \mapsto \liminf\left\{f_n(x) : n\in\mathbb{N}ight\}$
ewline
				Follows similarly.
		\end{itemize}
	
\end{enumerate}

Exercise 0.24 (Extended-real measurable functions)

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Exercise 0.24 (Extended-real measurable functions)

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