The Lebesgue integral · Qual Corpus

For measurable $f\ge0$, $\int f=\sup\{\int\varphi:\varphi\text{ simple},\ 0\le\varphi\le f\}$. The monotone convergence theorem: if $0\le f_n\uparrow f$, then $\int f_n\to\int f$. Fatou’s lemma: for measurable $f_n\ge0$, $\int\liminf_nf_n\le\liminf_n\int f_n$. The dominated convergence theorem: if $f_n\to f$ almost everywhere and $\abs{f_n}\le g$ with $g$ integrable, then $\int\abs{f_n-f}\to0$. The sequence $n\chi_{(0,1/n)}$ converges to $0$ at every point while every integral is $1$, and it has no integrable dominating function. For $\sigma$-finite measure spaces, Tonelli’s theorem equates the iterated integrals of a nonnegative measurable function on the product, and Fubini’s theorem does the same for a function integrable on the product. Chebyshev’s inequality: $m(\{\abs{f}\ge\lambda\})\le\lambda^{-1}\int\abs{f}$ for $\lambda>0$. For integrable $f$ and every $\varepsilon>0$ there is $\delta>0$ such that $\int_E\abs{f}<\varepsilon$ whenever $m(E)<\delta$.

The class $L^+$ of nonnegative measurable functions (Tag D-BF5L2)

Let $(X,\mcm)$ be a measurable space. The class $L^+$, or $L^+(X,\mcm)$, is the set of measurable functions $f\colon X\to[0,\infty]$.

Lebesgue integral of a nonnegative function (Tag D-R4VKE)

Let $(X,\mcm,\mu)$ be a measure space. For a nonnegative simple function $s = \sum_{j=1}^n c_j \chi_{E_j}$ with $c_j\in[0,\infty)$ and $E_j\in\mcm$, put \[ \int_X s \dmu \coloneqq \sum_{j=1}^n c_j \mu(E_j), \] with the convention $0\cdot\infty=0$. For $f\in$ $L^+$, the Lebesgue integral of $f$ is \[ \int_X f \dmu \coloneqq \sup \theset{ \int_X s \dmu \suchthat s \text{ simple and } 0\leq s \leq f }. \]

Monotone convergence theorem (Tag T-5K3IO)

For an almost everywhere increasing sequence in $L^+$, the integral of the limit is the limit of the integrals.Suggest a slogan

Let $(X,\mcm,\mu)$ be a measure space, and let $f_n$ for $n\geq1$ and $f$ be functions in $L^+(X,\mcm)$ such that, for almost every $x\in X$, $f_n(x)\leq f_{n+1}(x)$ for all $n$ and $f_n(x)\to f(x)$. Then \[ \lim_{n\to\infty} \int_X f_n\dmu = \int_X f\dmu . \]

Fatou's lemma and reverse Fatou's lemma (Tag T-LDJNS)

$\int\liminf f_n\le\liminf\int f_n$; if $f_n\le g$ with $g\in L^1$, then $\limsup\int f_n\le\int\limsup f_n$.Suggest a slogan

Let $(X,\mcm,\mu)$ be a measure space and let $f_n\colon X\to[0,\infty]$ be measurable for $n\geq1$.

(Fatou’s lemma [@Fol13]) \[ \liminf_{n\to\infty} \int_X f_n\dmu \geq \int_X \liminf_{n\to\infty} f_n\dmu . \]
(Reverse Fatou’s lemma) If moreover there is an integrable $g\colon X\to[0,\infty)$ with $f_n\leq g$ for all $n$, then \[ \limsup_{n\to\infty} \int_X f_n\dmu \leq \int_X \limsup_{n\to\infty} f_n\dmu . \]

The domination hypothesis in (2) cannot be dropped. On $\RR$ with Lebesgue measure, $f_n\coloneqq\chi_{[n,n+1]}$ satisfies $\int f_n=1$ for all $n$, while $\limsup_n f_n=0$ pointwise, so $\limsup_n\int f_n=1>0=\int\limsup_n f_n$.

Dominated convergence theorem (Tag T-IJQQG)

Almost-everywhere convergence under one integrable majorant gives $L^1$ convergence and convergence of integrals.Suggest a slogan

Let $(X,\mcm,\mu)$ be a measure space, and let $f_n\colon X\to\CC$ for $n\geq1$ and $f\colon X\to\CC$ be measurable with $f_n\to f$ almost everywhere. Suppose there is an integrable $g\colon X\to[0,\infty)$ with $\abs{f_n} \leq g$ almost everywhere for every $n$. Then $f$ is integrable and \[ \lim_{n\to\infty}\int_X \abs{f_n - f}\dmu = 0 ; \] consequently \[ \lim_{n\to\infty} \int_X f_n\dmu = \int_X f\dmu . \]

The functions $f_n$ may be complex-valued and of arbitrary sign; the only sign condition is $g\geq0$ on the dominating function.

Markov and Chebyshev inequalities (Tag PR-YO7MV)

Large-value sets are controlled by moments: $\mu\{|f|>\alpha\}\leq \alpha^{-p}\norm{f}_p^p$.Suggest a slogan

Let $(X,\mcm,\mu)$ be a measure space, let $f\colon X\to\CC$ be measurable, let $\alpha>0$, and put $S_\alpha\coloneqq\theset{x\in X\suchthat\abs{f(x)}>\alpha}$. Then \[ \mu(S_\alpha) \leq \frac{1}{\alpha}\int_X\abs{f}\dmu = \frac{1}{\alpha}\norm{f}_1, \] and more generally, for $0<p<\infty$, \[ \mu(S_\alpha) \leq \frac{1}{\alpha^p}\int_X\abs{f}^p\dmu = \frac{1}{\alpha^p}\norm{f}_p^p. \]

On $S_\alpha$ we have $\abs{f(x)} > \alpha$, hence $\abs{f(x)}^p > \alpha^p$, so \[ \int_X \abs{f}^p\dmu \geq \int_{S_\alpha} \abs{f}^p\dmu \geq \int_{S_\alpha} \alpha^p\dmu = \alpha^p \mu(S_\alpha). \] Dividing by $\alpha^p > 0$ gives the second inequality, and $p=1$ gives the first.

For a probability space $(\Omega,\mathcal F,P)$ and a nonnegative random variable $Y$ with expectation $E(Y)$, the case $p=1$ reads $P(Y>\alpha)\leq E(Y)/\alpha$.

Small tails of $L^1(\RR^d)$ functions (Tag T-RIY2P)

For $f\in L^1(\RR^d)$, $\int_{\abs{x}\ge r}\abs{f}\to0$ as $r\to\infty$.Suggest a slogan

Let $d\geq1$, let $m$ be Lebesgue measure on $\RR^d$, and let $f\colon\RR^d\to\CC$ be integrable with respect to $m$. For $r>0$, let $B_r(0)\coloneqq\theset{x\in\RR^d\suchthat\abs{x}<r}$. Then \[ \lim_{r\to\infty}\int_{\RR^d\setminus B_r(0)}\abs{f}\,dm=0. \] Equivalently, for every $\varepsilon>0$ there exists $r>0$ such that \[ \int_{\RR^d\setminus B_r(0)}\abs{f}\,dm<\varepsilon. \]

Tonelli's theorem for nonnegative measurable functions (Tag T-6PRW3)

Nonnegative measurable functions may be integrated iteratively in either order without an integrability hypothesis.Suggest a slogan

Let $f\colon\RR^n\times\RR^m\to[0,\infty]$ be Lebesgue measurable on $\RR^{n+m}$. For $x\in\RR^n$ let $f_x\colon\RR^m\to[0,\infty]$ be the slice $f_x(y)\coloneqq f(x,y)$. Then:

for almost every $x\in\RR^n$, the slice $f_x$ is Lebesgue measurable on $\RR^m$;
the function $F(x)\coloneqq\int_{\RR^m} f(x,y)\,dy\in[0,\infty]$, defined for almost every $x$, is Lebesgue measurable on $\RR^n$;
\[ \int_{\RR^{n+m}} f = \int_{\RR^n} F(x)\,dx = \int_{\RR^n}\qty{\int_{\RR^m} f(x,y)\,dy}dx \in[0,\infty], \] and the same holds with the roles of $x$ and $y$ exchanged [@Fol13].

Moreover, if $E\subseteq\RR^{n}\times\RR^m$ is Lebesgue measurable, then for almost every $x\in\RR^n$ the slice $E_x\coloneqq\theset{y\in\RR^m \suchthat (x, y) \in E}$ is Lebesgue measurable.

Fubini's theorem for integrable functions (Tag T-4GPEF)

Absolute integrability permits iterated integration in either order and recovers the integral on the product.Suggest a slogan

Let $f\colon\RR^n\times\RR^m\to\CC$ be integrable with respect to Lebesgue measure on $\RR^{n+m}$. For $x\in\RR^n$ let $f_x\colon\RR^m\to\CC$ be the slice $f_x(y)\coloneqq f(x,y)$. Then:

for almost every $x\in\RR^n$, the slice $f_x$ is integrable on $\RR^m$;
the function $F(x)\coloneqq\int_{\RR^m} f(x,y)\,dy$, defined for almost every $x$, is integrable on $\RR^n$;
\[ \int_{\RR^{n+m}} f = \int_{\RR^n}\qty{\int_{\RR^m} f(x,y)\,dy}dx , \] and the same holds with the roles of $x$ and $y$ exchanged.

Moreover, if $E\subseteq\RR^{n}\times\RR^m$ is Lebesgue measurable, then for almost every $x\in\RR^n$ the slice $E_x\coloneqq\theset{y\in\RR^m \suchthat (x, y) \in E}$ is Lebesgue measurable.

The hypothesis is $f$ integrable on $\RR^{n+m}$; $f$ may take values of both signs. Tonelli's theorem instead assumes $f\geq0$ measurable, with no integrability hypothesis.

State Fubini's theorem.

Fubini--Tonelli theorem (Tag T-X7XZX)

On a product of $\sigma$-finite spaces, if one iterated integral of $\abs{f}$ is finite, then $f\in L^1$ and both iterated integrals of $f$ equal $\int f$.Suggest a slogan

Let $(X,\mcm,\mu)$ and $(Y,\mcn,\nu)$ be $\sigma$-finite measure spaces, and let $f\colon X\times Y\to\CC$ be $\mcm\otimes\mcn$-measurable. Suppose that one of the iterated integrals of $\abs{f}$ is finite: \[ \int_X\qty{\int_Y\abs{f(x,y)}\,d\nu(y)}d\mu(x)<\infty\quad\text{or}\quad\int_Y\qty{\int_X\abs{f(x,y)}\,d\mu(x)}d\nu(y)<\infty. \] Then $f\in L^1(X\times Y,\mu\times\nu)$, and \[ \int_{X\times Y}f\,d(\mu\times\nu)=\int_X\qty{\int_Y f(x,y)\,d\nu(y)}d\mu(x)=\int_Y\qty{\int_X f(x,y)\,d\mu(x)}d\nu(y), \] where the inner integrals exist for almost every value of the outer variable.

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