Measure and measurable functions

Lebesgue outer measure $m^*$ is defined on every subset of $\RR$ and is countably subadditive. A set $E$ is measurable if $m^*(A)=m^*(A\cap E)+m^*(A\setminus E)$ for every $A\subseteq\RR$ (Carathéodory’s condition); the measurable sets form a $\sigma$-algebra on which $m^*$ is countably additive. A set $E$ is measurable if and only if for every $\varepsilon>0$ there is an open $U\supseteq E$ with $m^*(U\setminus E)<\varepsilon$, if and only if for every $\varepsilon>0$ there is a closed $F\subseteq E$ with $m^*(E\setminus F)<\varepsilon$. A function $f\colon E\to[-\infty,\infty]$ is measurable if $\{x:f(x)>a\}$ is measurable for every $a\in\RR$; suprema, infima, and pointwise limits of sequences of measurable functions are measurable. Egorov’s theorem: if $m(E)<\infty$ and measurable $f_n$ converge almost everywhere on $E$ to a finite-valued $f$, then for every $\varepsilon>0$ the convergence is uniform off a set of measure less than $\varepsilon$. Lusin’s theorem: if $m(E)<\infty$ and $f$ is measurable and finite almost everywhere on $E$, then for every $\varepsilon>0$ there is a closed $F\subseteq E$ with $m(E\setminus F)<\varepsilon$ and $f|_F$ continuous. The Borel–Cantelli lemma: if $\sum_nm(E_n)<\infty$, then almost every $x$ lies in only finitely many $E_n$.

Lebesgue outer measure (Tag D-3XE77)

A closed cube in $\RR^n$ is a product $Q=\prod_{i=1}^n[a_i,a_i+\ell]$ with $\ell\geq 0$, and its volume is $\abs{Q}\coloneqq\ell^n$. The outer measure of a set $E\subseteq\RR^n$ is \[ m_*(E) \coloneqq \inf \theset{\sum_{i=1}^\infty \abs{Q_i} \suchthat Q_1, Q_2, \ldots \text{ closed cubes with } E\subseteq\bigcup_{i=1}^\infty Q_i} \in [0,\infty]. \]

Lebesgue measurable sets (Tag D-MDJII)

Let $m_*$ be the outer measure on $\RR^n$. A subset $E\subseteq \RR^n$ is Lebesgue measurable if for every $\varepsilon> 0$ there exists an open set $O \supseteq E$ such that $m_*(O\setminus E) < \varepsilon$. For a Lebesgue measurable set $E$, its Lebesgue measure is $m(E) \coloneqq m_*(E)$.

Approximation of measurable sets by open, closed, and compact sets (Tag T-IIKSW)

Measurable sets are arbitrarily close in measure to open supersets and closed subsets, and finite-measure ones to compact subsets.Suggest a slogan

Let $E\subseteq\RR^n$ be Lebesgue measurable and let $\varepsilon>0$.

There exists an open set $O\supseteq E$ with $m(O\setminus E) < \varepsilon$.
There exists a closed set $F\subseteq E$ with $m(E\setminus F) < \varepsilon$.
If $m(E)<\infty$, there exists a compact set $K\subseteq E$ with $m(E\setminus K) < \varepsilon$ [@Fol13].

The finiteness hypothesis in (3) cannot be dropped. For $E=\RR^n$, every compact $K\subseteq\RR^n$ is bounded, hence has finite measure, so $m(\RR^n\setminus K)=\infty$.

Existence of a non-measurable subset of $\RR$ (Tag T-KZNWM)

Not every subset of the real line is Lebesgue measurable.Suggest a slogan

There exists a set $A\subseteq \RR$ that is not Lebesgue measurable.

Continuity of measure from below and from above (Tag PR-KKJ6O)

For increasing $E_n$, $\mu(E_n)\to\mu(\bigcup E_n)$; for decreasing $E_n$ with $\mu(E_1)<\infty$, $\mu(E_n)\to\mu(\bigcap E_n)$.Suggest a slogan

Let $(X,\mcm,\mu)$ be a measure space and let $E_n\in\mcm$ for $n\geq1$ [@Fol13].

If $E_1\subseteq E_2\subseteq\cdots$ and $E\coloneqq\bigcup_{n\geq1}E_n$, then $\mu(E_n)\to\mu(E)$.
If $E_1\supseteq E_2\supseteq\cdots$, $\mu(E_1)<\infty$, and $E\coloneqq\bigcap_{n\geq1}E_n$, then $\mu(E_n)\to\mu(E)$.

The hypothesis $\mu(E_1)<\infty$ in (b) is needed: for Lebesgue measure $m$ on $\RR$ and $E_n\coloneqq[n,\infty)$, $m(E_n)=\infty$ for all $n$ but $\bigcap_n E_n=\emptyset$.

State continuity of measure from above and from below.

Measurable function (Tag D-DHFN4)

Let $(X, \mcm_X)$ and $(Y, \mcm_Y)$ be measurable spaces. A function $f\colon X \to Y$ is $(\mcm_X, \mcm_Y)$-measurable if $\inverseof{f}(B)\in\mcm_X$ for every $B\in\mcm_Y$. Let $\mcb_\RR$ and $\mcb_{\RR^d}$ denote the Borel $\sigma$-algebras and $\mcl_{\RR^d}$ the $\sigma$-algebra of Lebesgue measurable subsets of $\RR^d$.

A function $f\colon X\to \RR$ on a measurable space $(X, \mcm_X)$ is measurable if it is $(\mcm_X, \mcb_\RR)$-measurable.
A function $f\colon \RR^d\to \RR$ is Borel measurable if it is $(\mcb_{\RR^d}, \mcb_\RR)$-measurable.
A function $f\colon \RR^d\to \RR$ is Lebesgue measurable if it is $(\mcl_{\RR^d}, \mcb_\RR)$-measurable.

Let $(X, \mcm_X)$ and $(Y, \mcm_Y)$ be measurable spaces, and let $\mce\subseteq\mcm_Y$ be a family of sets that generates $\mcm_Y$ as a $\sigma$-algebra. A function $f\colon X\to Y$ is $(\mcm_X,\mcm_Y)$-measurable if and only if $\inverseof{f}(E)\in\mcm_X$ for every $E\in\mce$.

In particular, since $\mcb_\RR$ is generated by each of the families of rays below, a function $f\colon X\to\RR$ is measurable if and only if any one of the following holds for every $\alpha \in \RR$:

$\inverseof{f}((\alpha, \infty))\in \mcm_X$;
$\inverseof{f}([\alpha, \infty))\in \mcm_X$;
$\inverseof{f}((-\infty, \alpha))\in \mcm_X$;
$\inverseof{f}((-\infty, \alpha])\in \mcm_X$.

The forward implication holds because $\mce\subseteq\mcm_Y$. Conversely, the family $\theset{B\subseteq Y \suchthat \inverseof{f}(B)\in\mcm_X}$ is a $\sigma$-algebra, because preimages commute with complements and countable unions, and it contains $\mce$; hence it contains the $\sigma$-algebra $\mcm_Y$ generated by $\mce$.

Simple function (Tag D-553MO)

Let $(X,\mcm)$ be a measurable space. A simple function on $X$ is a function $s\colon X\to\CC$ of the form \[ s(x) = \sum_{j=1}^n c_j \chi_{E_j}(x), \] where $n\geq 1$, $c_1,\ldots,c_n\in\CC$, $E_1,\ldots,E_n\in\mcm$, and $\chi_E$ is the indicator function of $E$.

Borel--Cantelli lemma (Tag T-OTR5M)

If $\sum_k m(E_k)<\infty$, then almost every point lies in only finitely many of the sets $E_k$.Suggest a slogan

Let $(E_k)_{k\geq1}$ be a sequence of Lebesgue measurable subsets of $\RR^n$ with \[ \sum_{k=1}^\infty m(E_k) < \infty . \] Then $m\qty{\limsup_{k\to\infty} E_k} = 0$, where $\limsup_k E_k$ is the limit superior of the sets $E_k$; equivalently, almost every $x\in\RR^n$ lies in only finitely many of the sets $E_k$.

The same statement holds for a sequence of measurable sets in any measure space. For a probability measure $P$ and events $E_k$ with $\sum_k P(E_k)<\infty$, it says that with probability $0$ infinitely many of the events $E_k$ occur.

Egorov's theorem (Tag FT-OGS76)

On a finite-measure set, almost-everywhere convergence becomes uniform after discarding arbitrarily little measure.Suggest a slogan

Let $m$ be Lebesgue measure on $\RR^n$, let $E\subseteq \RR^n$ be Lebesgue measurable with $m(E) < \infty$, and let $f_n\colon E\to\RR$ for $n\geq1$ and $f\colon E\to\RR$ be measurable with $f_n \to f$ almost everywhere on $E$. Then for every $\varepsilon > 0$ there exists a closed set $F\subseteq E$ such that $m(E\setminus F)<\varepsilon$ and $f_n\to f$ uniformly on $F$.

State Egorov's theorem.

Lusin's theorem (Tag T-CGFCU)

A measurable function on a set of finite measure has a continuous restriction to a closed subset whose complement has arbitrarily small measure.Suggest a slogan

Let $E\subseteq\RR^n$ be Lebesgue measurable with $m(E)<\infty$, and let $f\colon E\to\RR$ be measurable. For every $\varepsilon>0$ there exists a closed set $F_\varepsilon\subseteq\RR^n$ with \[ F_\varepsilon \subseteq E \quad\text{and}\quad m(E \setminus F_\varepsilon) \leq \varepsilon \] such that the restriction $\ro{f}{F_\varepsilon}\colon F_\varepsilon\to\RR$ is continuous.

Continuity of the restriction $\ro{f}{F_\varepsilon}$ does not make $f$ continuous at points of $F_\varepsilon$. Let $E=[0,1]$ and $f=\chi_{\QQ\cap[0,1]}$, which is discontinuous at every point of $[0,1]$. Enumerate $\QQ\cap[0,1]=\theset{q_1,q_2,\ldots}$ and put $F_\varepsilon\coloneqq[0,1]\setminus\bigcup_{k\geq1}(q_k-\varepsilon2^{-k-1},q_k+\varepsilon2^{-k-1})$. Then $F_\varepsilon$ is closed, $F_\varepsilon\subseteq E$, $m(E\setminus F_\varepsilon)\leq\varepsilon$, and $\ro{f}{F_\varepsilon}=0$ is continuous.

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