5 Measurable Function

measurable function

• $\mathrm{Let}(X,\mathcal{A})$ and $(Y,\mathcal{B})$ be measurable spaces. A map $f:X\to Y$ is called measurable if

$$B\in \mathcal{B}\Rightarrow f^{-1}(B)\in \mathcal{A}$$

• Let $(X,\mathcal{A})$ be a measurable space. A function $f:X\to \overline{\mathbb{R}}$ is measurable if it is measurable with respect to the Borel $\sigma$-algebra on $\overline{\mathbb{R}}$ associated to the standard topology as defined above.
• Let $X$ and $Y$ be topological spaces. A map $f:X\to Y$ is called Borel measurable if the pre-image of every Borel measurable subset of $Y$ under $f$ is a Borel measurable subset of $X$.

Remark. $\overline{\mathbb{R}}$ is defined as $\mathbb{R}\cup \{\infty \}\cup \{-\infty \}$. A subset of $\overline{\mathbb{R}}$ is open, if it is a countable union of the set of the form: $[-\infty ,a),(a,\infty ],(a,b)$.

Example. Define ${\chi }_A:X\to \mathbb{R}$ as ${\chi }_A(x)=1$ if $x\in A$; ${\chi }_A(x)=0$ if $x\notin A$. Then ${\chi }_A$ is measurable iff $A$ is a measurable set.

Proposition

Let $(X,\mathcal{A}),(Y,\mathcal{B}),(Z,\mathcal{C})$ be measurable spaces.

• The identity map ${\mathrm{id}}_X:X\to X$ is measurable.
• If $f:X\to Y$ and $g:Y\to Z$ are measurable maps, then so is the composition $g\circ f:X\to Z$.
• Let $f:X\to Y$ be any map. Then the set

$$f_{\ast }\mathcal{A}:=\left\{B\subset Y\mid f^{-1}(B)\in \mathcal{A}\right\}$$

is a $\sigma$-algebra on $Y$, called the pushforward of $\mathcal{A}$ under $f$.

• A map $f:X\to Y$ is measurable if and only if $\mathcal{B}\subset f_{\ast }\mathcal{A}$.

\begin{proof} Easy. \end{proof}

measurable and continuous maps

Let $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ be measurable spaces. Assume that $\mathcal{B}$ is the Borel $\sigma$-algebra of a topology on $Y$.

• A map $f:X\to Y$ is measurable if and only if for every open subset $U\subset Y$, $f^{-1}(U)\in \mathcal{A}$.
• Assume that $\mathcal{A}=\mathcal{B}(X)$ is the Borel $\sigma$-algebra of a topology on $X$. Then every continuous map $f:X\to Y$ is (Borel) measurable.

\begin{proof} Easy. \end{proof}

characterization of measurable functions

Let $(X,\mathcal{A})$ be $a$ measurable space and let $f:X\to \overline{\mathbb{R}}$ be any function. Then the following are equivalent:

• $f$ is measurable;
• $f^{-1}((a,\infty ])$ is a measurable subset of $X$ for every $a\in \mathbb{R}$;
• $f^{-1}([a,\infty ])$ is a measurable subset of $X$ for every $a\in \mathbb{R}$;
• $f^{-1}([-\infty ,b))$ is a measurable subset of $X$ for every $b\in \mathbb{R}$;
• $f^{-1}([-\infty ,b])$ is a measurable subset of $X$ for every $b\in \mathbb{R}$.

\begin{proof} Easy. \end{proof}

Lemma

Any open subset in ${\mathbb{R}}^n$ is a countable union of the sets in the form of $(a_1,b_1)\times \cdots \times (a_n,b_n)$.

\begin{proof} Let $O$ be the open subset in ${\mathbb{R}}^n$. Define $X:=\{x\in O:x=(x_1,\cdots ,x_n),x_i\in \mathbb{Q}\}$. Then $X$ is a countable set. For each $x\in X$, define $R_x=(a_1,b_1)\times \cdots \times (a_n,b_n)$ be the maximal cube such that $R_x\subseteq O$. Then ${\cup }_{x\in X}{R}_x=O$. \end{proof}

vector valued measurable functions

Let $(X,\mathcal{A})$ be a measurable space and let $f=\left(f_1,\cdots ,f_n\right):X\to {\mathbb{R}}^n$ be a function. Then $f$ is measurable if and only if $f_i:X\to \mathbb{R}$ is measurable for each $i$.

\begin{proof} Note that $f_i={\pi }_i\circ f$ and ${\pi }_i$ is continuous, then $f_i$ is measurable. Now assume that $f_i$ is measurable for each $i$. Let $Q(a,b)=(a_1,b_1)\times \cdots \times (a_n,b_n)$. Then

$$f^{-1}(Q(a,b))=\cap \{x\in X:f_i(x)\in (a_i,b_i)\}=\cap f_i^{-1}((a_i,b_i))\in \mathcal{A}.$$

Since any open subset in ${\mathbb{R}}^n$ is a countable union of the set $Q(a,b)$ by ^2bca2c, then $f$ is measurable. \end{proof}

Proposition

Let $(X,\mathcal{A})$ be a measurable space and let $u,v:X\to \mathbb{R}$ be measurable functions. If $\varphi :{\mathbb{R}}^2\to \mathbb{R}$ is continuous, then the function $h:X\to \mathbb{R}$ defined by

$$h(x)=\varphi (u(x),v(x)),x\in X$$

is measurable.

\begin{proof} Since $f:=(u,v)$ and $\varphi$ are measurable, then $h$ is measurable. \end{proof}

properties of measurable functions

Let $(X,\mathcal{A})$ be a measurable space.

• If $f,g:X\to \mathbb{R}$ are measurable functions, then so are the functions

$$f+g;fg;\max \{f,g\};\min \{f,g\};|f|.$$

• Let $f_k:X\to \overline{\mathbb{R}},k=1,2,\dots$ be a sequence of measurable functions. Then the following functions from $X$ to $\overline{\mathbb{R}}$ are measurable:

$$\inf f_k,\sup f_k,\lim \sup f_k,\lim \inf f_k.$$

\begin{proof} i) By ^12414f.

ii) Define $g:={\sup }_k{f}_k:X\to \overline{\mathbb{R}}$ and let $a\in \mathbb{R}$. Then the set

$$\begin{array}{rl}{g}^{-1}((a,\infty ])& =\left\{x\in X:\underset{i}{\sup }{f}_i(x)>a\right\}=\left\{x\in X:\exists i\text{ such that }{f}_i(x)>a\right\}\\ & =\bigcup _{i=1}^{\infty }\left\{x\in X\mid f_i(x)>a\right\}=\bigcup _{i=1}^{\infty }{f}_i^{-1}((a,\infty ]).\end{array}$$

is measurable. Hence it follows that $g$ is measurable. It also follows that $-f_k$ is measurable, and hence the function $\inf f_k=-\sup \left(-f_k\right)$ is measurable. Also we conclude that the functions

$$\underset{k}{\mathrm{limsup}}{f}_k=\underset{l\in \mathbb{N}}{\inf }\underset{k\ge l}{\sup }{f}_k;\underset{k}{\mathrm{liminf}}{f}_k=\underset{l\in \mathbb{N}}{\sup }\underset{k\ge l}{\inf }{f}_k$$

are also measurable. \end{proof}

Corollary

If $f_n:X\to \overline{\mathbb{R}}$ is measurable and $f_n(x)\to f(x)$ for all $x$, then $f:X\to \mathbb{R}$ is measurable.

\begin{proof} By ^e11178, ii). \end{proof}

Proposition

If $f:\mathbb{R}\to \mathbb{R}$ is monotone, then $f$ is Borel measurable.

\begin{proof} Suppose that $f$ is increasing, for otherwise let $g=-f$. For any $a\in \mathbb{R}$, define $x_0=\sup \{x:f(x)⩽a\}$.

• If $f(x_0)>a$, then $f^{-1}((a,\infty ))=[x_0,\infty )$.
• If $f(x_0)⩽a$, then $f^{-1}((a,\infty ))=(x_0,\infty )$.

Therefore, $f^{-1}((a,\infty ))$ is measurable and so $f$ is Borel measurable. \end{proof}