Exercise 1.4.29 (Properties of measurable functions)

Question

Let $(X,\mathcal{B})$ be a measurable space. Show that\begin{enumerate}[label=(oman*)]
\item a function $f:X	o[0,+\infty]$ is measurable if and only if the level sets $\{x\in X: f(x)>\lambda\}$ are $\mathcal{B}$-measurable.\item the indicator function $1_E:X	o\{0,1\}$ is measurable if and only if $E \subseteq X$ is $\mathcal{B}$-measurable.\item a function $f:X	o\mathbb{C}$ is measurable if and only if for every Borel measurable subset $E\subseteq \mathbb{C}$ the set $f^{-1}(E)$ is $\mathcal{B}$-measurable.\item a function $f:X	o\mathbb{C}$ is measurable if and only if its real and imaginary parts are measurable.\item a function $f:X	o\mathbb{R}$ is measurable if and only if its positive and negative magnitudes are measurable.\item if $\left(fight)_{n\in\mathbb{N}}:X	o\mathbb{C}$ is a sequence of measurable functions converging to another function $f:X	o\mathbb{C}$, then $f$ is also measurable.\item if $f:X	o\mathbb{C}$ is measurable, and $\phi:\mathbb{C}	o\mathbb{C}$ is continuous, then $\phi \circ f$ is measurable.\item the sum of product of two measurable functions is measurable.\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(oman*)]
	\item If $f$ is measurable, then the sets $\{x\in X: f(x)>\lambda\} = f^{-1}((\lambda,+\infty])$ must be measurable too since the latter is the inverse image of an open set in $[0,+\infty]$.
ewline
	Conversely, suppose that for any $\lambda>0$ the set $f^{-1}((\lambda,+\infty])$ is $\mathcal{B}$-measurable. Then also      \begin{itemize}\item sets of the form $f^{-1}([-\infty,\lambda]) = f^{-1}(\mathbb{R}/(\lambda,+\infty]) = X/f^{-1}((\lambda,+\infty])$ must be measurable.
	
	\item sets of the form $f^{-1}([-\infty,\lambda)) = f^{-1}(\bigcap_{n=1}^{\infty}[-\infty,\lambda+1/n]) = \bigcap_{n=1}^{\infty}f^{-1}([-\infty,\lambda+1/n])$ must be measurable.
	
	\item sets of the form $f^{-1}(a,b) = f^{-1}(a,+\infty] \cap f^{-1}[-\infty,b)$ are measurable.\end{itemize}      Pick an arbitrary open set $U \subseteq [0,+\infty]$. Then $U$ is a countable union of open intervals $\bigcup_{n\in\mathbb{N}}I_n$ (cf. Analysis I). \href{https://math.stackexchange.com/questions/2692405/every-open-set-in-the-extended-real-line-overline-mathbb-r-is-a-countable}{Recall} that an open interval of the extended positive real line either contains $+\infty$ (in which case it can be written as $(a,+\infty]$ or $[0,+\infty]$ for $a \in [0,+\infty]$) or it does not (in which case it can be written as $(a,b)$ for $a,b\in [0,+\infty]$). Both of the cases result in measurable inverse images under $f$ by the previous argument. We then have $f^{-1}\left(\bigcup_{n\in\mathbb{N}}I_night) = \bigcup_{n\in\mathbb{N}}f^{-1}(I_n)$, and the last term is measurable as a countable union of measurable sets.\item Suppose the $1_E$ is measurable. Then by definition
	$$E = (1_E)^{-1}(\{1\}) = (1_E)^{-1}\left(\bigcap_{n=1}^{\infty}\left(-\frac{1}{n}+1,1+\frac{1}{n}ight)ight) = \bigcap_{n=1}^{\infty} (1_E)^{-1} \left(-\frac{1}{n}+1,1+\frac{1}{n}ight)$$
	and the last term is measurable as a countable intersection of measurable sets.
ewlineNow suppose conversely that $E$ is measurable. Now pick an arbitrary open set $U\subseteq [0,+\infty]$. Notice that only the values $0,1\in[0,+\infty]$ get mapped into. Thus, we have four cases:      \begin{itemize}\item $0\in U, 1
otin U$, then $f^{-1}(U) = f^{-1}(\{0\}) \cap f^{-1}(U/\{0\}) = X/E \cup \emptyset = X/E$ which is measurable.\item $0 
otin U, 1 \in U$. Then $f^{-1}(U) = f^{-1}(\{1\}) \cap f^{-1}(U/\{1\}) = E \cup \emptyset = E$ which is measurable.\item $0,1
otin U$ or $0,1\in U$, in both cases the inverse image $f^{-1}(U)$ is measurable as it is either $\emptyset$ or $X$ respectively.\end{itemize}\item Suppose that the inverse image of every Borel measurable set in $\mathbb{C}$ is $\mathcal{B}$-measurable. Then, in particular the inverse image of every open set must me $\mathcal{B}$-measurable too, and we are done.
ewlineNow suppose conversely that $f$ is measurable. We use the $\sigma$-induction from the Remark 1.4.15.
	\begin{itemize}
		\item $f^{-1}(\emptyset) = \emptyset$ is $\mathcal{B}$-measurable.
		\item If $f^{-1}(E)$ is $\mathcal{B}$-measurable, then $f^{-1}(\mathbb{C}/E) = X/f^{-1}(E)$ is $\mathcal{B}$-measurable, too.\item If $f^{-1}(E_n)$ is $\mathcal{B}$-measurable for $\left(Eight)_{n\in\mathbb{N}}$, then $f^{-1}(\bigcup_{n=1}^{\infty}E_n) = \bigcup_{n=1}^{\infty}f^{-1}(E_n)$ is $\mathcal{B}$-measurable, too.
		\item Finally, the property holds for every element of the generating set $\mathcal{O}(\mathbb{C})$ of open sets.
	\end{itemize}
	Thus, this property must hold for all elements of $\sigma(\mathcal{O}(\mathbb{C}))$ too.
	
	\item Suppose that $f = [\Re f, \Im f]$ is measurable. We verify that the real part $\Re f:X	o\mathbb{R}$ is measurable. Pick an arbitrary open set $U \subseteq \mathbb{R}$ (for the real part). Then the inverse image is $(\Re f)^{-1}(U) =  f^{-1}(U 	imes \mathbb{R})$. The latter is a $\mathcal{B}$-measurable set since it is an inverse image of a product of two open sets which is open (cf. Topology). Similarly the imaginary part is measurable as well.
ewlineNow suppose that the real and the imaginary part of $f$ are measurable. Pick an arbitrary open set $U\subseteq \mathbb{C}$. Let $c\in U$. Notice that      $$f^{-1}(c) = (\Re f)^{-1}(\Re c) \cap (\Im f)^{-1}(\Im c) = (f^{-1}\Re^{-1})(c) \cap (f^{-1}\Im^{-1})(c) $$      This is easy to verify. (If $x\in f^{-1}(c)$ then $f(x)=c$ and therefore $\Re f (x) = \Re c$ hence $f(x) \in \Re^{-1}(\Re c)$ and $x \in f^{-1}\Re^{-1}(\Re c)$, similarly with the imaginary part. Conversely, $x\in f^{-1}\Re^{-1}(\Re c)$ means $\Re f (x) = \Re c$ and $x \in f^{-1}\Im^{-1}(\Im c)$ means $\Im f(x) = \Im c$. Since $f(x) = \Re f(x) + \Im f(x) = c$, the result follows.) But both terms are $\mathcal{B}$-measurable by assumption, and so is their intersection.\item Let $f:X	o\mathbb{R}$ be measurable, and consider $f^+,f^-:X	o[0,+\infty)$ given by $f^+ = \max\{f,0\}$ and $f^-=\min\{f,0\}$. Then any open set $U\subseteq [0,+\infty]$ we have $(f^+)^{-1}(U) = f^{-1}(U)$ is $\mathcal{B}$-measurable, and $(f^-)^{-1}(U) = f^{-1}(-U)$ which is also $\mathcal{B}$-measurable, since $-U$ is open.
ewlineNow suppose that $f^+,f^-:X	o[0,+\infty)$ given by $f^+ = \max\{f,0\}$ and $f^-=\min\{f,0\}$ are measurable. Pick an arbitrary open set $U\subseteq\mathbb{R}$. We than have
	\begin{align*}
	f^{-1}(U) 
	&= f^{-1}([-\infty,0] \cap U)\cup f^{-1}(U \cap [0,+\infty])\
	&=(-f^-)^{-1}([-\infty,0] \cap U)\cup (f^+)^{-1}(U \cap [0,+\infty])\
	&=(f^-)^{-1}(-U\cap [0,+\infty])\cup (f^+)^{-1}(U \cap [0,+\infty])
	\end{align*}
	Both sets are open in $[0,+\infty]$, and the union of $\mathcal{B}$-measurable inverses is again $\mathcal{B}$-measurable.\item We argue that $f^{-1}(E)$ is a countable union/intersection of $\mathcal{B}$-measurable sets. In particular,
	$$f^{-1}(U) = \bigcup_{n=1}^{\infty}\bigcap_{N=n}^{\infty} f^{-1}_N(U).$$
	\begin{itemize}
		\item[$\subseteq$)] Pick an arbitrary $x \in f^{-1}(U)$. Then $f(x)\in U$ is an interion point, and we thus can find an $\epsilon>0$ such that $B(f(x),\epsilon)\subseteq U$. Since $\left(fight)_{n\in\mathbb{N}}	o f$, we can find a $n\in\mathbb{N}$ such that $\forall N>n$ we have $f_N(x) \in B(f(x),\epsilon)$ and thus $f_N(x)\in U$. This implies $x \in \bigcup_{n=1}^{\infty}\bigcap_{N=n}^{\infty} f^{-1}_N(U)$
		\item[$\supseteq$)] Pick an arbitrary $x \in  \bigcup_{n=1}^{\infty}\bigcap_{N=n}^{\infty} f^{-1}_N(U)$, i.e., there exists a $n'\in\mathbb{N}$ such that $x\in \bigcap_{N=n'}^{\infty} f^{-1}_N(U)$, i.e., for all $N\geq n$ we have $f_N(x)\in U$. Furthermore, there exists a $n''\in\mathbb{N}$ such that for all $N\geq n''$ we have $d(f_N(x),f(x)) \leq \epsilon$. Thus, after $n=\max\{n',n''\}$ the values $f_N(x)$ are both $\epsilon$-close to $f(x)$ and contained in $U$. By triangle inequality $f(x)$ must be contained in an $\epsilon$-ball in $U$ as well.
		\end{itemize}
		
	\item The inverse image $(\phi \circ f)^{-1}(U) = f^{-1}\phi^{-1}(U)$ is measurable, since $U$ and therefore also $\phi^{-1}(U)$ are open, and $f$ is measurable.\item Let $f,g:X	o\mathbb{C}$ be measurable. Consider two further functions
	\begin{align*}
	l:X	o\mathbb{C}^2, \quad &x \mapsto [f(x),g(x)]\
	\phi:\mathbb{C}^2	o\mathbb{C},\quad &[c_1,c_2] \mapsto c_1 + c_2
	\end{align*}
	Then $l$ is measurable. To see why, pick an arbitrary open $U \subseteq \mathbb{C}^2$. Then $l^{-1}(U) = (\pi_1 l)^{-1}(\pi_1 U) \cap (\pi_2 l)^{-1}(\pi_2 U)$ where $\pi_i$ is the projection function in the $i$-th co-ordinate. But $\pi_il$ is a composition of a continuous function with a measurable function, and is thus by part (vii) itself measurable. Taking the intersection of $\mathcal{B}$-measurable sets we see that $l$ is measurable.
ewlineFurthermore $\phi$ is continuous (cf. \href{../analysis_II}{Analysis II}). Thus, by part (vii) the function $f+g = l \circ \phi$ is measurable.
ewline
	A similar argument proves that the product is measurable as well.
\end{enumerate}

Exercise 1.4.29 (Properties of measurable functions)

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Exercise 1.4.29 (Properties of measurable functions)

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