Integration – Notes

remark (A common proof technique):

Show something holds for indicator functions.
Show it holds for simple functions by linearity.
Use $s_k \nearrow f$ and apply MCT to show it holds for $f$ .

remark (on notation):

$L^+$ : nonnegative measurable functions
$L^1$ : Lebesgue integrable functions, so ${\left\lVert {f} \right\rVert}_{L^1} \coloneqq\int {\left\lvert {f} \right\rvert} < \infty$ .

definition (Measurable Function):

A function $f: (X, {\mathcal{M}}_X) \to (Y, {\mathcal{M}}_Y)$ is $({\mathcal{M}}_X, {\mathcal{M}}_Y){\hbox{-}}$ measurable iff $f^{-1}({\mathcal{M}}_Y) \subseteq {\mathcal{M}}_X$ . Equivalently, if ${\mathcal{E}}_Y$ is a generating set for ${\mathcal{B}}_Y$ , $f^{-1}({\mathcal{E}}_Y) \subseteq {\mathcal{B}}_X$ .

An functional on a general measurable space $f: f:(X, {\mathcal{M}}_X) \to ({\mathbf{R}}, {\mathcal{B}}_{\mathbf{R}})$ is measurable $\iff f$ is $({\mathcal{M}}_X, {\mathcal{B}}_{\mathbf{R}}){\hbox{-}}$ measurable.
A functional $f: {\mathbf{R}}^d\to {\mathbf{R}}$ is Borel measurable iff $f$ is $({\mathcal{B}}_{{\mathbf{R}}^d}, {\mathcal{B}}_{\mathbf{R}}){\hbox{-}}$ measurable.
A functional $f: {\mathbf{R}}^d\to {\mathbf{R}}$ is Lebesgue measurable iff $f$ is $({\mathcal{L}}_{{\mathbf{R}}^d}, {\mathcal{B}}_{\mathbf{R}}){\hbox{-}}$ measurable.

Using that ${\mathcal{B}}_{{\mathbf{R}}}$ is generated by open/closed rays, it suffices to check any of the following (for all $\alpha \in {\mathbf{R}}$ ):

$f^{-1}(\alpha, \infty)\in {\mathcal{M}}$
$f^{-1}[\alpha, \infty)\in {\mathcal{M}}$
$f^{-1}(-\infty, \alpha)\in {\mathcal{M}}$
$f^{-1}(-\infty, \alpha]\in {\mathcal{M}}$

remark:

Note that we still require Borel sets in the target for Lebesgue measurability! Taking $({\mathcal{L}}_{{\mathbf{R}}^d}, {\mathcal{L}}_{\mathbf{R}})$ functions is too stringent, e.g. this class does not contain continuous functionals.

warnings:

If $f$ is ${\mathcal{L}}{\hbox{-}}$ measurable and $h$ is continuous, it’s not necessarily true that $k\coloneqq f\circ h$ is ${\mathcal{L}}{\hbox{-}}$ measurable. Standard counterexample: set $g(x) \coloneqq C(x) + x$ for $C$ the Cantor-Lebesgue function, then $g:[0, 1]\to [0, 2]$ is a homeomorphism. Then $m(g(C)) = 1$ since $f$ is constant on intervals in $C^c$ , so use Vitali’s theorem: a set is null iff every subset is measurable. So $g(C)$ contains a non-measurable set $A$ . Define $B\coloneqq g^{-1}(A)$ , then $B \subset C$ and $m(C) = 0$ implies $B$ is measurable and $\chi_B$ is a measurable function. But then $k\coloneqq\chi_B \circ g^{-1}$ is not ${\mathcal{L}}{\hbox{-}}$ measurable, since $k^{-1}(1) = A$ is a non-measurable set, but $\chi_B$ is ${\mathcal{L}}{\hbox{-}}$ measurable and $g^{-1}$ is continuous.

proposition (Closure of measurable functions under operations):

${\mathcal{M}}{\hbox{-}}$ measurable functionals are closed under

Sums
Products
Sups/infs
Limsups/Liminfs
Limits when they exist, and the limiting function is measurable.
$\max(f, g)$ and $\min(f, g)$ .

Characteristic functions on measurable sets are automatically measurable, since $E\in {\mathcal{M}}\implies E = \chi_E^{-1}(\left\{{1}\right\})$ .

definition (Simple Function):

A simple function $s: {\mathbf{C}}\to X$ is a finite linear combination of indicator functions of measurable sets, i.e. $\begin{align*} s(x) = \sum_{j=1}^n c_j \chi_{E_j}(x) .\end{align*}$

proposition (Folland 2.10b):

If $f:X\to {\mathbf{C}}$ is measurable, there is a sequence of simple functions $\phi_n\nearrow f$ which always converges pointwise, and converges uniformly on any bounded set.

definition (Lebesgue Integral):

$\begin{align*} \int_X f \coloneqq\sup \left\{{ \int s(x) \,d\mu{~\mathrel{\Big\vert}~}0\leq s \leq f, s\text{ simple } }\right\} .\end{align*}$

Note that if $s = \sum c_j \chi_{E_j}$ is simple, then $\begin{align*} \int_X s(x) \,d\mu\coloneqq\sum_{j=1}^n c_j \mu(E_j) .\end{align*}$

remark (Integrals split across disjoint sets):

A useful fact: for $(X, \mathcal{M})$ a measure space, integrals split across disjoint sets: $\begin{align*} \int_X f = \int_{X\setminus A} f + \int_A f && \forall\, A \in \mathcal{M} .\end{align*}$

definition (Essential supremum and infimum, essentially bounded):

An essential lower bound $b$ on a function $f$ is any real number such that $S_{b} \coloneqq\left\{{x{~\mathrel{\Big\vert}~}f(x) < b }\right\} = f^{-1}(-\infty, b)$ has measure zero. The essential infimum is the supremum of all essential lower bounds, i.e. ${\mathrm{ess}}\inf f \coloneqq\sup_{b} \left\{{b{~\mathrel{\Big\vert}~}\mu S_b = 0}\right\}$ . This is the greatest lower bound almost everywhere.

Similarly an essential upper bound $c$ is any number such that $S^c \coloneqq f^{-1}(c, \infty)$ has measure zero, and the essential supremum is ${\mathrm{ess}}\sup f \coloneqq\inf_{c} \left\{{c{~\mathrel{\Big\vert}~}\mu S^c = 0}\right\}$ , which is the least upper bound almost everywhere.

A function is essentially bounded if ${\left\lVert {f} \right\rVert}_\infty \coloneqq{\mathrm{ess}}\sup f < \infty$ . These are functions which are bounded almost everywhere.

example (An essentially bounded but not bounded function):

$f(x) = x\chi_{\mathbf{Q}}(x)$ is essentially bounded but not bounded.

proposition (L-infty functions are equivalent to bounded almost-everywhere functions):

If $f\in L^\infty(X)$ , then $f$ is equal to some bounded function $g$ almost everywhere.

theorem (p-Test for Integrals):

$\begin{align*} \int_0^1 {1\over x^p} < \infty \iff p < 1 \\ \int_1^\infty {1\over x^p} < \infty \iff p > 1 .\end{align*}$

slogan:

Large powers of $x$ help us in neighborhoods of infinity and hurt around zero.

The Convergence Theorems

theorem (Monotone Convergence):

If $f_n: X\to [0, \infty) \in L^+$ and $f_n \nearrow f$ almost everywhere, then $\begin{align*} \lim \int f_n = \int \lim f_n = \int f \quad \text{i.e.}~~ \int f_n \to \int f .\end{align*}$

slogan:

Measurable, non-negative, increasing pointwise a.e. allows commuting limits and integrals.

#todo Proof

theorem (Dominated Convergence):

If $f_n \in L^1$ and $f_n \to f$ almost everywhere with ${\left\lvert {f_n} \right\rvert} \leq g$ for some $g\in L^1$ , then $f\in L^1$ and $\begin{align*} \int {\left\lvert {f_n - f} \right\rvert} \to 0 .\end{align*}$

As a consequence, $\begin{align*} \lim \int f_n = \int \lim f_n = \int f \quad \text{i.e.}~~ \int f_n \to \int f < \infty \end{align*}$

Positivity not needed.

#todo Proof

theorem (Generalized DCT):

$f_n \in L^1$ with $f_n \to f$ almost everywhere,
There exist $g_n\geq 0 \in L^1$ nonnegative with ${\left\lvert {f_n} \right\rvert} \leq g_n$ ,
$g_n\to g$ almost everywhere with $g\in L^1$ , and
$\lim \int g_n = \int g$ ,

then $f\in L^1$ and $\lim \int f_n = \int f < \infty$ .

Note that this is the DCT with ${\left\lvert {f_n} \right\rvert} < {\left\lvert {g} \right\rvert}$ relaxed to ${\left\lvert {f_n} \right\rvert} < g_n \to g\in L^1$ .

proof (of generalized DCT):

Proceed by showing $\limsup \int f_n \leq \int f \leq \liminf \int f_n$ :

$\int f \geq \limsup \int f_n$ : $\begin{align*} \int g - \int f &= \int \qty{g-f} \\ &\leq \liminf \int \qty{g_n - f_n} \quad \text{Fatou} \\ &= \lim \int g_n + \liminf \int (-f_n) \\ &= \lim \int g_n - \limsup \int f_n \\ &= \int g - \limsup \int f_n \\ \\ \implies \int f &\geq \limsup \int f_n .\end{align*}$
- Here we use $g_n - f_n \overset{n\to\infty}\longrightarrow g-f$ with $0 \leq {\left\lvert {f_n} \right\rvert} - f_n \leq g_n - f_n$ , so $g_n - f_n$ are nonnegative (and measurable) and Fatou’s lemma applies.
$\int f \leq \liminf \int f_n$ : $\begin{align*} \int g + \int f &= \int(g+f) \\ &\leq \liminf \int \qty{g_n + f_n} \\ &= \lim \int g_n + \liminf \int f_n \\ &= \int g + \liminf f_n \\ \\ \int f &\leq \liminf \int f_n .\end{align*}$
- Here we use that $g_n + f_n \to g+f$ with $0 \leq {\left\lvert {f_n} \right\rvert} + f_n \leq g_n + f_n$ so Fatou’s lemma again applies.

remark:

The converse to the DCT does not hold, i.e. $L^p$ boundedness does not imply a.e. boundedness, i.e. it is not true that $\lim \int f_k = \int f$ implies that $\exists g\in L^p$ such that $f_k < g$ a.e. for every $k$ .

Take

$b_k = \sum_{j=1}^k \frac 1 j \to \infty$
$f_k = \chi_{[b_k, b_{k+1}]}$

Then

$f_k \overset{a.e.}\to f = 0$ ,
$\int f_k = \frac 1 k \to 0 \implies {\left\lVert {f_k} \right\rVert}_p \to 0$ ,
$0 = \int f = \lim \int f_k = 0$
But $g > f_k \implies g > {\left\lVert {f_k} \right\rVert}_\infty = 1$ a.e. $\implies g\not\in L^p({\mathbf{R}})$ .

proposition (Convergence in

$L^1$ implies convergence of norms):

If $f\in L^1$ , then $\begin{align*} {\left\lVert {f_n - f} \right\rVert}_{L^1} \overset{n\to\infty}\longrightarrow 0 \iff {\left\lVert {f_n} \right\rVert}_{L^1} \overset{n\to\infty}\longrightarrow {\left\lVert {f} \right\rVert}_{L^1} .\end{align*}$

proof (That

$L^1$ convergence implies convergence of norms):

Let $g_n = {\left\lvert {f_n} \right\rvert} - {\left\lvert {f_n - f} \right\rvert}$ , then $g_n \to {\left\lvert {f} \right\rvert}$ and $\begin{align*} {\left\lvert {g_n} \right\rvert} = {\left\lvert { {\left\lvert {f_n} \right\rvert} - {\left\lvert {f_n - f} \right\rvert} } \right\rvert} \geq {\left\lvert {f_n - (f_n - f)} \right\rvert} = {\left\lvert {f} \right\rvert} \in L^1 ,\end{align*}$ so the DCT applies to $g_n$ and $\begin{align*} {\left\lVert {f_n - f} \right\rVert}_1 = \int {\left\lvert {f_n - f} \right\rvert} + {\left\lvert {f_n} \right\rvert} - {\left\lvert {f_n} \right\rvert} = \int {\left\lvert {f_n} \right\rvert} - g_n\\ \to_{DCT} \lim \int {\left\lvert {f_n} \right\rvert} - \int {\left\lvert {f} \right\rvert} .\end{align*}$

theorem (Fatou):

If $f_n$ is a sequence of nonnegative measurable functions, then $\begin{align*} \liminf_n \int f_n &\geq \int \liminf_n f_n \\ \limsup_n \int f_n &\leq \int \limsup_n f_n .\end{align*}$

#todo Prove

example (Using Fatou to compute the limit of a sequence of integrals):

$\begin{align*} \lim _{n \rightarrow \infty} \int_{0}^{\infty} \frac{n^{2}}{1+n^{2} x^{2}} e^{-\frac{x^{2}}{n^{3}}} d x \overset{\text{Fatou}}\geq \int_{0}^{\infty} \lim _{n \rightarrow \infty} \frac{n^{2}}{1+n^{2} x^{2}} e^{-\frac{x^{2}}{n^{3}}} d x \to \int \infty .\end{align*}$

Note that MCT might work, but showing that this is non-decreasing in $n$ is difficult.

Commuting

proposition (Commuting Sums with Integrals):

$\begin{align*} f_n \geq 0 \text{ and } \sum_n \int {\left\lvert {f_n} \right\rvert} = \sum_n {\left\lVert {f_n} \right\rVert}_{L^1} < \infty \implies \sum_n \int f_n = \int \sum_n f_n .\end{align*}$ If the $f_n$ are not necessarily non-negative, we still have $\begin{align*} \left\{{f_n}\right\} \subseteq L^1 \text { and }\qty{\sum\int{\left\lvert {f_n} \right\rvert} < \infty \text { or } \int \sum {\left\lvert {f_n} \right\rvert} < \infty } \implies \int\sum_n f_n = \sum_n \int f_n .\end{align*}$

proof (Commuting sums with integrals, non-negative case):

Idea: MCT.
Let $F_N = \sum^N f_n$ be a finite partial sum;
Then there are simple functions $\phi_n \nearrow f_n$
So $\sum^N \phi_n \nearrow F_N$ and MCT applies

proof (Commuting sums with integrals, integrable case):

By Tonelli, if $f_n(x) \geq 0$ for all $n$ , taking the counting measure allows interchanging the order of “integration”.
By Fubini on ${\left\lvert {f_n} \right\rvert}$ , if either “iterated integral” is finite then the result follows.

proposition (?):

$\begin{align*} \left\{{f_n}\right\} \subseteq L^1 \text{ and } \sum_n {\left\lVert {f_n} \right\rVert}_{L^1} < \infty \implies \sum_n f_n \text{ converges }{ \text{a.e.} }\text{ and in } L^1 .\end{align*}$

proof (?):

Define $F_N = \sum^N f_k$ and $F = \lim_N F_N$ , then ${\left\lVert {F_N} \right\rVert}_1 \leq \sum^N {\left\lVert {f_k} \right\rVert} < \infty$ so $F\in L^1$ and ${\left\lVert {F_N - F} \right\rVert}_1 \to 0$ so the sum converges in $L^1$ . Almost everywhere convergence: ?

proposition (Commuting derivatives with integrals, Folland 2.27):

If $f:X\times I \to {\mathbf{C}}$ where $f_t: X\to {\mathbf{C}}$ is integrable for each $t$ , then if ${\left\lvert {f(x, t)} \right\rvert} \leq {\left\lvert {g(x)} \right\rvert}$ for some $g\in L^1$ , then $\begin{align*} \lim_{t\to t_0}\int_X f(x, t) \,d\mu= \int_X f(x, t_0) \,d\mu\coloneqq F(t_0) ,\end{align*}$ and if $f_x: I\to {\mathbf{C}}$ is continuous for all $x$ , then $F: I\to {\mathbf{C}}$ is continuous.

Moreover if ${\frac{\partial f}{\partial t}\,}$ exists and ${\left\lvert {{\frac{\partial f}{\partial t}\,}(x, t)} \right\rvert} \leq {\left\lvert {g} \right\rvert}$ for some $g\in L^1$ , then $\begin{align*} {\frac{\partial }{\partial t}\,} \int_X f(x, t) \,d\mu = \int_X {\frac{\partial }{\partial t}\,} f(x, t) \,d\mu .\end{align*}$