guide:B9a63f187d

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Integration for positive (nonnegative) functions

In this chapter we will introduce the integral as a new concept in terms of a measure, which is a more general point of view instead of the Riemann integral, which has several drawbacks. Consider for example the function

[[math]] \one_\mathbb{Q}(x):=\begin{cases}1,&x\in\mathbb{Q}\\0,&x\not\in\mathbb{Q}\end{cases} [[/math]]

There is no way where we can say that this function would be Riemann integrable and thus we cant make sense of the integral

[[math]] \int_0^1\one_\mathbb{Q}(x)dx [[/math]]

in terms of the theory of Riemann integration. However, with the notion of an integral with respect to a positive measure we can also deal with such integrals as we will see. Let, in this chapter, [math](E,\mathcal{A})[/math] be a measurable space and let [math]f(x)=\sum_{i=1}^n\alpha_i\one_{A_i}(x)[/math] be a simple function, where [math]A_i\in \A[/math] and [math]\alpha_i\in\R[/math] for all [math]1\leq i\leq n[/math] and [math]n\in\N[/math]. Moreover, let us assume without loss of generality that

[[math]] \alpha_1 \lt \alpha_2 \lt ... \lt \alpha_n. [[/math]]

Then [math]A_i=f^{-1}(\{\alpha_i\})\in\mathcal{A}[/math]. Let us also denote by [math]\mu[/math] a positive measure on [math](E,\A)[/math]. Then we can define the integral for simple functions as follows.

Definition (Integral with respect to a measure)

Assume that [math]f[/math] takes values in [math]\mathbb{R}_+[/math] with [math]0\leq \alpha_1 \lt ... \lt \alpha_n[/math]. Then the Integral of [math]f[/math] with respect to [math]\mu[/math] is defined by

[[math]] \int f d\mu=\sum_{i=1}^{n}\alpha_i\mu(A_i), [[/math]]

where we use the convention [math]0\cdot \infty=0[/math], in case that [math]\alpha_i=0[/math] and [math]\mu(A_i)=\infty[/math]. Moreover, if [math]f(x)=\one_{A}(x)+0\cdot \one_{A^C}(x)[/math], for [math]A\in\A[/math], then

[[math]] \int f d\mu=\mu(A)(+0\cdot \mu(A^C)). [[/math]]

It is easy to obtain that by definition, if [math]\int fd\mu\in[0,\infty][/math], then if [math]f=0[/math] we get that [math]\int fd\mu=0[/math].

This integral is well defined. Indeed, let

[[math]] f(x)=\sum_{i=1}^n\alpha_i\one_{A_i}(x) [[/math]]

be the canonical form of the simple function of [math]f[/math], i.e. [math]\alpha_i[/math] distinct, [math]A_i=f^{-1}(\{\alpha_i\})[/math]. Then, by definition, it follows that

[[math]] \int fd\mu =\sum_{i=1}^n\alpha_i\mu(A_i). [[/math]]

On the other hand, we can also write [math]f[/math] as

[[math]] f(x)=\sum_{j=1}^m\beta_j\one_{B_j}(x), [[/math]]

where [math](\beta_j)_{1\leq j\leq m}[/math] forms an [math]\mathcal{A}[/math]-partition, with [math]\beta_j\geq 0[/math] and where the [math]\beta_j[/math]'s are not necessarily distinct. We want to show that the integral is still the same. Note that for each [math]i\in\{1,...,n\}[/math], we get that [math]A_i[/math] is the disjoint union of the sets [math]B_j[/math] for which [math]\alpha_i=\beta_j[/math]. Then the additivity property of the measure shows that

[[math]] \mu(A_i)=\sum_{\{j\mid \beta_j=\alpha_j\}}\mu(B_j), [[/math]]

and therefore the integral doesn't depend on the representation of [math]f[/math].

Proposition

Let [math]f[/math] and [math]g[/math] be two simple, positive and measurable functions on [math]E[/math]. Then

• For [math]a,b\geq 0[/math] we get
  [[math]] \int (af+bg)d\mu=a\int fd\mu+b\int gd\mu. [[/math]]
• If [math]f\leq g[/math], then
  [[math]] \int fd\mu\leq \int gd\mu. [[/math]]

Show Proof

For [math](i)[/math], let us first set

[[math]] f=\sum_{i=1}^n\alpha_i\one_{A_i}g=\sum_{k=1}^m\alpha_k'\one_{A_k'}. [[/math]]

Moreover, we can note that

[[math]] A_i=\bigcup_{k=1}^m(A_i\cap A_k'),A_k'=\bigcup_{i=1}^n(A_k'\cap A_i) [[/math]]

and therefore we can write

[[math]] f=\sum_{j=1}^p\beta_j\one_{B_j}g=\sum_{j=1}^p\gamma_j\one_{B_j}, [[/math]]

where [math](B_j)_{1\leq j\leq pj}[/math] is an [math]\mathcal{A}[/math]-partition of [math]E[/math] obtained from a reordering of [math](A_i\cap A_k')[/math]. Thus we get

[[math]] \int fd\mu=\sum_{j=1}^p\beta_j\mu(B_j),\int gd\mu=\sum_{j=1}^p\gamma_j\mu(B_j). [[/math]]

It follows that

[[math]] \int (af+bg)d\mu=\sum_{j=1}^p(a\beta_j+b\gamma_j)\mu(B_j)=a\sum_{j=1}^p\beta_j\mu(B_j)+b\sum_{j=1}^p\gamma_j\mu(B_j)=a \int f d\mu+b\int gd\mu. [[/math]]

For [math](ii)[/math], we can write [math]g[/math] as [math]g=f+g-f[/math], where [math]f\geq 0[/math] and [math]g-f\geq 0.[/math] Then

[[math]] \int g d\mu=\underbrace{\int(g-f)d\mu}_{\geq 0}+\int fd\mu\geq \int fd\mu. [[/math]]

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Definition (Integral for measurable maps)

Let [math](E,\mathcal{A},\mu)[/math] be a measure space. Moreover, let us denote by [math]\mathcal{E}^+[/math] the set of all nonnegative simple functions and let [math]f:E\longrightarrow [0,\infty)[/math] be a measurable map. Then we can define the integral for [math]f[/math] to be given as

[[math]] \int fd\mu:=\sup_{h\leq f\atop h\in\mathcal{E}^+}\int hd\mu. [[/math]]

Sometimes we have different notation for the same.

[[math]] \int fd\mu=\int f(x)d\mu(x)=\int f(x)\mu(dx). [[/math]]

Proposition

Let [math](E,\A,\mu)[/math] be a measure space and let [math]f,g:E\longrightarrow [0,\infty)[/math] be measurable maps. Then

• If [math]f\leq g[/math], then
  [[math]] \int fd\mu\leq \int gd\mu. [[/math]]
• If [math]\mu(\{x\in E\mid f(x) \gt 0\})=0,[/math] then
  [[math]] \int fd\mu=0. [[/math]]

Show Proof

Exercise.

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Theorem (Monotone convergence theorem)

Let [math](f_n)_{n\in\N}[/math] be an increasing sequence of positive and measurable functions (with values in [math][0,\infty)[/math]), and let [math]f=\lim_{n\to \infty}\uparrow f_n[/math]. Then

[[math]] \int fd\mu=\lim_{n\to\infty}\int f_nd\mu. [[/math]]

Show Proof

We know that since [math]f_n\leq f[/math], we have [math]\int f_nd\mu\leq \int f d\mu[/math] and hence

[[math]] \lim_{n\to\infty}\int f_nd\mu\leq \int fd\mu. [[/math]]

We need to show that

[[math]] \int fd\mu\leq \lim_{n\to\infty}\int f_nd\mu. [[/math]]

Let therefore [math]h=\sum_{i=1}^{m}\alpha_i\one_{A_i}[/math], such that [math]h\leq f[/math], and let [math]a\in[0,1)[/math]. Moreover let then

[[math]] E_n:=\{x\in E\mid ah(x)\leq f_n(x)\}. [[/math]]

We see immediately that [math]E_n[/math] is measurable and since [math]f_n\uparrow f[/math] as [math]n\to\infty[/math] and [math]a \lt 1[/math], we see that

[[math]] E=\bigcup_{n\geq 1}E_n,E_n\subset{E_{n+1}}. [[/math]]

We also note that [math]f_n\geq a\one_{E_n}h[/math]. To see this, wee need to emphasize two cases.

• If [math]x\in E_n[/math], then [math]a\one_{E_n}(x)h(x)=ah(x)[/math] and by definition of [math]E_n[/math] we get that[math]f_n(x)\geq ah(x).[/math]
• If [math]x\not\in E_n[/math], then [math]a\one_{E_n}(x)h(x)=0[/math] and thus [math]f_n(x)\geq 0[/math] holds since [math]f_n[/math] is positive.

Hence it follows that

[[math]] \int f_nd\mu\geq \int a\one_{E_n}hd\mu=a\int\one_{E_n}hd\mu=a\sum_{i=1}^m\alpha_i\mu(A_i\cap E_n). [[/math]]

Since [math]E_n\uparrow E[/math] and [math]A_i\cap E_n\uparrow A_i[/math] as [math]n\to\infty[/math], we have [math]\mu(A_i\cap E_n)\uparrow \mu(A_i)[/math] as [math]n\to\infty[/math]. Thus we get that

[[math]] \lim_{n\to\infty}\int f_nd\mu\geq a\sum_{i=1}^m\alpha_i\mu(A_i)=a\int hd\mu. [[/math]]

Note that the left side doesn't depend on [math]a[/math]. Now we let [math]a\to 1[/math] and obtain then

[[math]] \lim_{n\to\infty}\int f_nd\mu\geq \int hd\mu. [[/math]]

This is now true for every [math]h\in\mathcal{E}^+[/math] with [math]h\leq f[/math] and the left hand side doesn't depend on [math]h[/math]. Therefore we have

[[math]] \lim_{n\to\infty}\uparrow \int f_nd\mu\geq \sup_{h\in\mathcal{E}^+\atop h\leq f}\int hd\mu=\int fd\mu. [[/math]]

Let us recall that for any positive measurable map [math]f[/math] (values in [math][0,\infty)[/math]) there exists an increasing sequence [math](f_n)_{n\in\N}[/math] of simple positive functions such that [math]f=\lim_{n\to\infty}f_n[/math].

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Proposition

Let [math](E,\A,\mu)[/math] be a measure space. Then

• If [math]f[/math] and [math]g[/math] are two positive and measurable on [math]E[/math] and [math]a,b\in\mathbb{R}_+[/math], then
  [[math]] \int (af+bg)d\mu=a\int fd\mu + b\int gd\mu. [[/math]]
• If [math](f_n)_{n\in\N}[/math] is a sequence of measurable and positive functions on [math]E[/math], then
  [[math]] \int\sum_n f_nd\mu=\sum_n\int f_nd\mu. [[/math]]

Show Proof

For [math](i)[/math], take two positive sequences [math](f_n)_{n\in\N}[/math] and [math](g_n)_{n\in\N}\geq 0[/math] of simple functions such that [math]f_n\uparrow f[/math] and [math]g_n\uparrow g[/math] as [math]n\to\infty[/math]. Then we get

[[math]] \int (af+bg)d\mu\stackrel{MCV}{=}\lim_{n\to\infty}\int (af_n+bg_n)d\mu=\lim_{n\to\infty}\left(a\int f_n d\mu+b\int g_nd\mu\right)\stackrel{MCV}{=}a\int fd\mu +b\int gd\mu, [[/math]]

where [math]MCV[/math] stands for monotone convergence. Now [math](ii)[/math] is an Immediate consequence of the monotone convergence theorem (MCV). Indeed, set

[[math]] g_N=\sum_{n=1}^Nf_n, [[/math]]

with [math]g_N\geq 0[/math] and [math]\lim_{N\to\infty}\uparrow g_N=\sum_{n=1}^\infty f_n[/math]. If we now apply the monotone convergence theorem to [math](g_N)_{N\in\N}[/math], we get

[[math]] \begin{multline*} \int\sum_{n=1}^{\infty}f_n d\mu=\int \lim_{N\to\infty}g_Nd\mu \stackrel{MCV}{=}\lim_{N\to\infty}\int g_Nd\mu\\=\lim_{N\to\infty}\int\sum_{n=1}^Nf_nd\mu\stackrel{(i)}{=}\lim_{N\to\infty}\sum_{n=1}^N\int f_nd\mu=\sum_{n=1}^{\infty}\int f_n d\mu, \end{multline*} [[/math]]

which proves the proposition.

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Example

Let us consider the Dirac measure [math]\delta_x[/math] for [math]x\in E[/math] and let [math]f:E\longrightarrow \mathbb{R}_+[/math] be a measurable map. Then

[[math]] \int fd\delta_x=f(x). [[/math]]

Example

Let us consider the space [math]\mathbb{N}[/math], the [math]\sigma[/math]-Algebra [math]\mathcal{P}(\mathbb{N})[/math] and the counting measure on [math]\N[/math], which is the measure [math]\mu[/math] satisfying that for all [math]A\subset \N[/math] we get [math]\mu(A)=\vert A\vert[/math]. Then for a measurable map [math]f:\mathbb{N}\longrightarrow \mathbb{R}_+[/math] we get

[[math]] \int fd\mu=\sum_{n\in\mathbb{N}}f(n). [[/math]]

Moreover, we also get for every positive sequence [math](a_{n,k})_{n\in\mathbb{N},k\in\mathbb{N}}\geq 0[/math] that

[[math]] \sum_{k\in\mathbb{N}}\sum_{n\in\mathbb{N}}a_{n,k}=\sum_{n\in\mathbb{N}}\sum_{k\in\mathbb{N}}a_{n,k}. [[/math]]

Corollary

Let [math](E,\A)[/math] be a measurable space and let [math]f[/math] be a positive measurable map on [math]E[/math]. Moreover, let us define for [math]A\in\mathcal{A}[/math] a map [math]\nu[/math] on [math]E[/math] by

[[math]] \nu(A):=\int_{E}\one_{A}(x)f(x)d\mu=\int_Af(x)d\mu. [[/math]]

Then [math]\nu[/math] is a measure on [math](E,\mathcal{A})[/math], which is called the measure with density [math]f[/math] with respect to [math]\mu[/math] and we write

[[math]] \nu=f\circ \mu. [[/math]]

It is clear that if [math]\mu(A)=0[/math] then [math]\nu(A)=0[/math] for all [math]A\in\A[/math].

Show Proof

First of all [math]\nu(\varnothing)=0[/math] follows from the given proposition 1.4. Now let [math](A_n)_{n\in\mathbb{N}}\in\mathcal{A}[/math] be a sequence of measurable sets with [math]A_n\cap A_m=\varnothing[/math] for all [math]n\not=m[/math]. Then

[[math]] \nu\left(\bigcup_{n\geq 1}A_n\right)=\int\one_{\bigcup_{n\geq 1}A_n}(x)f(x)d\mu=\int\sum_{n\geq 1}\one_{A_n}(x)f(x)d\mu=\sum_{n\geq 1}\int \one_{A_n}(x)f(x)d\mu=\sum_{n\geq 1}\nu(A_n), [[/math]]

which actually shows that [math]\nu[/math] is a measure on [math]E[/math].

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We say that a property is true [math]\mu[/math]-almost everywhere and we write [math]\mu[/math]-a.e. (or a.e. if it is clear for which measure), if this property holds on a set [math]A\in\mathcal{A}[/math] with [math]\mu(A^C)=0[/math]. For example, if [math]f[/math] and [math]g[/math] are both measurable maps on [math]E[/math], then [math]f=g[/math] a.e. means that

[[math]] \mu(\{x\in E\mid f(x)\not=g(x)\})=0. [[/math]]

Proposition

Let [math](E,\A,\mu)[/math] be a measure space and Let [math]f,g:E\longrightarrow \R[/math] be measurable and positive functions. Then the following hold.

• [math]\mu(\{x\in E\mid f(x) \gt a\})\leq \frac{1}{a}\int fd\mu[/math] for all [math]a \gt 0[/math].
• If [math]\int fd\mu \lt \infty[/math], then [math]f \lt \infty[/math] a.e.
• [math]\int fd\mu=0[/math] if and only if [math]f=0[/math] a.e.
• If [math]f=g[/math] a.e., then [math]\int fd\mu=\int gd\mu[/math].

Show Proof

We show each points seperately.

• Consider the set [math]A_a=\{x\in E\mid f(x) \gt a\}[/math]. Then we can observe that [math]f(x)\geq a\one_{A_a}(x)[/math] for all [math]x\in A_a[/math]. Then it follows that
  [[math]] \int f(x)d\mu\geq \int a\one_{A_a}(x)d\mu [[/math]]
  and
  [[math]] \mu(A_a)\leq \frac{1}{a}\int fd\mu. [[/math]]
• For [math]n\geq 1[/math] consider the sets [math]A_n=\{x\in E\mid f(x)\geq n\}[/math] and [math]A_\infty=\{x\in E\mid f(x)=\infty\}=\bigcap_{n\in\N}A_n[/math]. Then we get that
  [[math]] \mu(A_\infty)=\lim_{n\to\infty}\mu(A_n), [[/math]]
  which implies that
  [[math]] \mu(A_n)\leq \underbrace{\frac{1}{n}\int f(x)d\mu}_{\xrightarrow{n\to\infty} 0} \lt \infty. [[/math]]
• We have already seen that if [math]f=0[/math] a.e., then [math]\int fd\mu=0[/math]. Therefore, it is enough to show the other direction. Let [math]n \gt 1[/math]. Then
  [[math]] \mu(B_n)\leq n\underbrace{\int fd\mu}_{0}=0. [[/math]]
  Thus we get that
  [[math]] \mu(\{x\in E\mid f(x) \gt 0\})=\mu\left(\bigcup_{n\geq 1}\right)\leq \sum_{n\geq 1}\mu(B_n)=0. [[/math]]
• Let us introduce a special notation at this point. We write [math]f\lor g[/math] for [math]\sup(f,g)[/math] and [math]f\land g[/math] for [math]\inf(f,g)[/math]. Now assume that [math]f=g[/math] a.e., which implies that [math]f\lor g=f\land g[/math] and hence we get
  [[math]] \int (f\lor g)d\mu=\int (f\land g)d\mu+\int (f\lor g-f\land g)d\mu=\int (f\land g)d\mu. [[/math]]
  Because of the fact that
  [[math]] \begin{align*} f\land g\leq f\leq f\lor g&\Longrightarrow \int fd\mu=\int(f\lor g)d\mu=\int (f\land g)d\mu\\ f\land g\leq g\leq f\lor g&\Longrightarrow \int gd\mu=\int (f\lor g)d\mu=\int (f\land g)d\mu, \end{align*} [[/math]]
  we finally get
  [[math]] \int fd\mu=\int gd\mu. [[/math]]

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Theorem (Fatou's lemma)

Let [math](f_n)_{n\in\N}[/math] be a sequence of real valued, measurable and positive functions on a measure space [math](E,\A,\mu)[/math]. Then

[[math]] \int \liminf_{n\to\infty}f_nd\mu\leq \liminf_{n\to\infty}\int f_nd\mu. [[/math]]

Show Proof

Recall that we actually have

[[math]] \liminf_{n\to\infty} f_n=\lim_{n\to\infty}\nearrow \left(\inf_{k\geq n}f_n\right),\left(=\sup_n\inf_{k\geq n}f_k\right). [[/math]]

From the monotone convergence theorem (MCV) we get

[[math]] \int \liminf_{n\to\infty} f_nd\mu\stackrel{MCV}{=}\lim_{n\to\infty}\int \inf_{k\geq n}f_kd\mu. [[/math]]

Now for all [math]p\geq k[/math] we have

[[math]] \begin{align*} \inf_{n\geq k}f_n &\leq f_p\\ \Longrightarrow\int \inf_{n\geq k}f_nd\mu &\leq \int f_pd\mu\\ \Longrightarrow\int \inf_{n\geq k}f_nd\mu &\leq \inf_{p\geq k}\int f_pd\mu\\ \Longrightarrow\lim_{k\to\infty}\int\inf_{n\geq k}f_nd\mu &\leq \lim_{k\to\infty}\uparrow\inf_{p\geq k}\int f_pd\mu=\liminf_{n\to\infty}\int f_nd\mu, \end{align*} [[/math]]

which proves the claim.

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Integrable functions

Definition (Integrable)

Let [math](E,\A,\mu)[/math] be a measure space and let [math]f:E\longrightarrow \R[/math] be a measurable map. We say that [math]f[/math] is integrable with respect to [math]\mu[/math] if

[[math]] \int \vert f\vert d\mu \lt \infty. [[/math]]

Moreover, if [math]f[/math] is integrable, we define its integral to be given by

[[math]] \int fd\mu =\int f^+d\mu -\int f^-d\mu. [[/math]]

It always holds that

[[math]] \int f^{\pm}d\mu\leq \int \vert f\vert d\mu. [[/math]]

For instance

[[math]] \int_{\mathbb{R}^+}\frac{\sin(x)}{x}dx=\frac{\pi}{2},\text{but}\int_{\mathbb{R}^+}\left\vert\frac{\sin(x)}{x}\right\vert dx=\infty. [[/math]]

Moreover, we will denote by [math]\mathcal{L}^1(E,\mathcal{A},\mu)[/math] the space of integrable (and measurable) functions. Furthermore, we denote by [math]\mathcal{L}^1_+(E,\mathcal{A},\mu)[/math] the same space, but containing only positive functions.

Proposition

Let [math](E,\A,\mu)[/math] be a measure space. Then the following hold.

• If [math]f\in\mathcal{L}^1(E,\mathcal{A},\mu)[/math], then [math]\left\vert\int fd\mu\right\vert \leq \int \vert f\vert d\mu.[/math]
• [math]\mathcal{L}^1(E,\mathcal{A},\mu)[/math] is a vector space and the map [math]f\mapsto \int\vert f\vert d\mu[/math] is a linear form.
• If [math]f,g\in\mathcal{L}^1(E,\mathcal{A},\mu)[/math] and [math]f\leq g[/math], then [math]\int fd\mu\leq \int gd\mu.[/math]
• If [math]f,g\in\mathcal{L}^1(E,\mathcal{A},\mu)[/math] and [math]f=g[/math] a.e., then [math]\int fd\mu=\int gd\mu.[/math]

Show Proof

We show each point seperately.

• Let [math]f\in\mathcal{L}^1(E,\mathcal{A},\mu)[/math]. Then we can obtain
  [[math]] \begin{align*} \left\vert \int fd\mu\right\vert &=\left\vert \int f^+d\mu-\int f^-d\mu\right\vert\\ &\leq \left\vert \int f^+d\mu\right\vert +\left\vert\int f^-d\mu\right\vert\\ &=\int f^+d\mu +\int f^- d\mu\\ &=\int \left(f^++f^-\right)d\mu\\ &=\int \vert f\vert d\mu. \end{align*} [[/math]]
• Indeed, [math]\mathcal{L}^1(E,\mathcal{A},\mu)[/math] is a linear space and for [math]f\in \mathcal{L}^1(E,\A,\mu)[/math] we get that the map
  [[math]] \begin{equation} \label{map} f\longmapsto\int fd\mu \end{equation} [[/math]]
  is a linear form. First we want to show that [math]\int afd\mu=a\int fd\mu[/math] for some [math]a\in\R[/math]. Let therefore [math]a\in \R[/math] and consider the case where [math]a\geq 0[/math] and the one where [math]a \lt 0[/math] as follows.
  [[math]] \begin{align*} \underline{\underline{a\geq 0}}:\int(af)d\mu&=\int(af)^+d\mu-\int (af)^-d\mu\\ &= a\int f^+d\mu -a\int f^-d\mu\\ &= a\int fd\mu. \end{align*} [[/math]]
  [[math]] \begin{align*} \underline{\underline{a \lt 0}}:\int (af)d\mu&=\int (af)^+d\mu-\int (af)^-d\mu\\ &=(-a)\int f^-d\mu -\int f^+d\mu\\ &=a\int fd\mu. \end{align*} [[/math]]
  If [math]f[/math] and [math]g[/math] are in [math]\mathcal{L}^1(E,\mathcal{A},\mu)[/math], the inequality [math]\vert f+g\vert\leq \vert f\vert +\vert g\vert[/math] implies that [math](f+g)\in\mathcal{L}^1(E,\mathcal{A},\mu)[/math]. One has to check the linearity of the map (map). It is easy to obtain the following implications.
  [[math]] \begin{align*} (f+g)^+-(f+g)^-&=(f+g)=f^+-f^-+g^+-g^-\\ \Longrightarrow(f+g)^++f^-+g^-&=(f+g)^-+f^++g^+\\ \Longrightarrow\int (f+g)^+d\mu+\int f^-d\mu+\int g^-d\mu&=\int (f+g)^-d\mu+\int f^+d\mu+\int g^+d\mu\\ \Longrightarrow\int (f+g)^+d\mu-\int(f+g)^-d\mu&=\int f^+d\mu-\int f^-d\mu+\int g^+d\mu-\int g^-d\mu\\ \Longrightarrow\int (f+g)d\mu&=\int fd\mu+\int gd\mu. \end{align*} [[/math]]
• Let [math]f,g\in\mathcal{L}^1(E,\mathcal{A},\mu)[/math] with [math]f\leq g[/math]. We can write [math]g=f+(g-f)[/math] and by assumption [math]g-f\geq 0[/math], which also implies that [math]\int (g-f)d\mu\geq 0[/math]. Therefore we have
  [[math]] \int gd\mu=\int fd\mu+\int(g-f)d\mu\geq \int f d\mu, [[/math]]
  which proves the claim.
• Let [math]f,g\in\mathcal{L}^1(E,\A,\mu)[/math] with [math]f=g[/math] a.e., which also implies that [math]f^+=g^+[/math] a.e. and [math]f^-=g^-[/math] a.e. Thus it follows that
  [[math]] \int f^+d\mu=\int g^+d\mu\text{and}\int f^-d\mu=\int g^-d\mu. [[/math]]
  Therefore [math]\int fd\mu=\int gd\mu[/math].

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Extension to the complex case

Let [math](E,\A,\mu)[/math] be a measure space and let [math]f:E\longrightarrow \C(\cong \mathbb{R}^2)[/math] be a measurable map, which basically means that [math]Re(f)[/math] and [math]Im(f)[/math] are both measurable maps. We say that [math]f[/math] is integrable if [math]Re(f)[/math] and [math]Im(f)[/math] are both integrable, or equivalently

[[math]] \int \vert f\vert d\mu \lt \infty, [[/math]]

and we write [math]f\in\mathcal{L}^1_\C(E,\A,\mu)[/math]. This is simply because of the fact that

[[math]] \int fd\mu=\int (Re(f)+Im(f))d\mu=\int Re(f)d\mu+\int Im(f)d\mu. [[/math]]

Moreover, the properties [math](i),(ii)[/math] and [math](iv)[/math] of proposition also hold for the complex case.

Lebesgue's dominated convergence theorem

An important question of integration theory is whether the interchanging of limits and integrals is actually possible and under which condition on the sequence of maps on some measure space. We have already seen the monotone convergence theorem and Fatou's lemma, giving some simple conditions for such an interchange. Another way of achieving the same with different conditions is due to Lebesgue, who gave a more general condition, which is going to be discussed now.

Theorem (Lebesgue's dominated convergence theorem)

Let [math](E,\A,\mu)[/math] be a measure space and let [math](f_n)_{n\in\N}[/math] be a sequence of functions in [math]\mathcal{L}^1(E,\mathcal{A},\mu)[/math] (resp. [math]\mathcal{L}^1_{\mathbb{C}}(E,\mathcal{A},\mu)[/math]). Moreover, assume that the following hold.

• There exists a measurable map [math]f[/math] on [math]E[/math] with values in [math]\mathbb{R}[/math] (resp. [math]\mathbb{C}[/math]) such that for all [math]x\in E[/math]
  [[math]] \lim_{n\to\infty}f_n(x)=f(x). [[/math]]
• There exists a psotive and measurable map [math]g:E\longrightarrow \mathbb{R}[/math] such that
  [[math]] \int gd\mu \lt \infty [[/math]]
  and such that [math]\vert f_n\vert \leq g[/math] a.e. for all [math]n\in\N[/math].

Then [math]f\in \mathcal{L}^1(E,\mathcal{A},\mu)[/math] (resp. [math]\mathcal{L}^1_{\mathbb{C}}(E,\mathcal{A},\mu))[/math] and we have

[[math]] \lim_{n\to\infty}\int\vert f_n-f\vert d\mu=0 [[/math]]

and hence

[[math]] \lim_{n\to\infty}\int f_nd\mu=\int fd\mu. [[/math]]

Show Proof

Let us first assume some stronger assumptions.

• [math]\lim_{n\to\infty}f_n(x)=f(x)[/math] for all [math]x\in E[/math].
• There exists a positive and measurable map [math]g:E\longrightarrow \mathbb{R}[/math] such that
  [[math]] \int gd\mu \lt \infty [[/math]]
  and [math]\vert f_n(x)\vert \geq g(x)[/math] for all [math]x\in E[/math].

Consider the general case where [math](i)[/math] and [math](ii)[/math] hold. Now define the set

[[math]] A=\left\{x\in E\mid f_n(x)\xrightarrow{n\to\infty}f(x)\text{and}\forall n, \vert f_n(x)\vert\leq g(x)\right\}. [[/math]]

Then [math]\mu(A^C)=0[/math]. Let us now apply the first part of the proof to

[[math]] \begin{align*} \tilde{f}_n(x)&=\one_{A}(x)f_n(x)\\ \tilde{f}(x)&=\one_{A}(x)f(x). \end{align*} [[/math]]

Thus we have that [math]f=\tilde{f}[/math] a.e. and [math]f_n=\tilde{f}_n[/math] a.e. Therefore, we get the following equations.

[[math]] \begin{align*} \int f_nd\mu&=\int \tilde f_nd\mu\\ \int fd\mu&=\int \tilde fd\mu\\ \lim_{n\to\infty}\int f_nd\mu&=\int fd\mu\\ \lim_{n\to\infty}\int\vert f-f_n\vert d\mu&=0 \end{align*} [[/math]]

Since now [math]\vert f\vert \leq g[/math] and [math]\int gd\mu \lt \infty[/math], we get that [math]\int \vert f\vert d\mu \lt \infty[/math]. Hence we have

[[math]] \vert f-f_n\vert \leq 2g\text{and}\vert f-f_n\vert\xrightarrow{n\to\infty}0. [[/math]]

Now, applying Fatou's lemma, we can observe that

[[math]] \liminf_{n\to\infty}\int 2g-\vert f-f_n\vert d\mu\geq \int \liminf_{n\to\infty} 2g-\vert f-f_n\vert d\mu, [[/math]]

and therefore

[[math]] \liminf_{n\to\infty} \int 2g-\vert f-f_n\vert d\mu\geq \int 2g d\mu, [[/math]]

which implies that

[[math]] \int 2gd\mu-\limsup_{n\to\infty}\int \vert f-f_n\vert d\mu\geq \int 2g d\mu. [[/math]]

it is easy now to observe that [math]\limsup_{n\to\infty}\int \vert f-f_n\vert d\mu\leq 0,[/math] which basically implies that [math]\lim_{n\to\infty}\int \vert f-f_n\vert d\mu =0[/math] and thus we can finally deduce

[[math]] \left\vert \int fd\mu-\int f_nd\mu\right\vert=\left\vert\int(f-f_n)d\mu\right\vert\leq \int \vert f-f_n\vert d\mu\xrightarrow{n\to\infty}0 [[/math]]

which simply means that

[[math]] \lim_{n\to\infty}\int f_n d\mu=\int fd\mu. [[/math]]

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Parameter Integrals

In this section we want to consider the special case of an integral. Basically, we want to look at integrals of the form

[[math]] \int f(u,x)d\mu, [[/math]]

for some [math]u\in E[/math], which actually gives rise to a map

[[math]] \begin{align*} F:E&\longrightarrow \R\\ u&\longmapsto F(u)=\int f(u,x)d\mu \end{align*} [[/math]]

Example

[Gamma function] Let us start with a special example of such a map. The Gamma function is defined as the map

[[math]] \begin{align*} \Gamma:\C&\longrightarrow \R\\ s&\longmapsto \Gamma(s)=\int_0^\infty t^{s-1}e^{-t}dt, \end{align*} [[/math]]

The question is, whether this integral converges for all [math]s\in\C[/math]. This is certainly not the case and only possible if [math]Re(s) \gt 0[/math]. An important functional equation is given by

[[math]] \Gamma(n+1)=n!, [[/math]]

where [math]n\in\N[/math]. We will need this function later to describe the volume of the unit ball in [math]\R^d[/math].

Theorem

Let [math](E,\A,\mu)[/math] be a measure space. Let [math](U,d)[/math] be a metric space and let [math]f:U\times E\longrightarrow\Big|_\C^\R[/math] with [math]u_0\in U[/math]. Moreover, assume that the following hold.

• The map [math]x\mapsto f(u,x)[/math] is measurable for all [math]x\in U[/math].
• The map [math]u\mapsto f(u,x)[/math] is continuous at [math]u_0[/math] a.e.
• There exists a measurable function [math]g\in\mathcal{L}^1(E,\mathcal{A},\mu)[/math] such that for all [math]u\in U[/math]
  [[math]] \vert f(u,x)\vert \leq g(x)\text{-a.e.} [[/math]]

Then the map [math]F(u)=\int_Ef(u,x)d\mu[/math] is well defined and continuous at [math]u_0[/math].

Show Proof

From [math](iii)[/math] follows that the map [math]x\mapsto f(u,x)[/math] is integrable for every [math]u\in U[/math], and so [math]F(u)[/math] is well defined. Take a sequence [math](u_n)_{n\in\N}\in U[/math] such that [math]u_n\xrightarrow{n\to \infty} u_0[/math], which basically means that [math]d(u_n,u_0)\xrightarrow{n\to \infty} 0[/math]. Then by continuity from [math](ii)[/math], we get that [math]f(u_n,x)\xrightarrow{n\to \infty} f(u_0,x)[/math] a.e. and from [math](iii)[/math] we can apply Lebesgue's dominated convergence theorem to obtain that

[[math]] \lim_{n\to\infty}F(u_n)=\lim_{n\to\infty}\int f(u_n,x)d\mu=\int f(u_0,x)d\mu=F(u_0). [[/math]]

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Observe that [math]F(u)[/math] is continuous if [math]F[/math] is continuous at every point [math]u\in U[/math].

Example

[Fourier analysis] we want to give several examples of Fourier analysis at this point.

• Let [math]\mu[/math] be a measure on [math](\mathbb{R},\mathcal{B}(\mathbb{R}))[/math] such that [math]\mu(\{x\})=0[/math] for all [math]x\in\R[/math]. Moreover, let [math]\phi\in\mathcal{L}^1(\R,\B(\R),\mu)[/math]. Then the map
  [[math]] F(u)=\int_\R\one_{(-\infty,u]}(x)\phi(x)d\mu=\int_{(-\infty,u]}\phi(x)d\mu [[/math]]
  is continuous. Here we have [math]f(u,x)=\one_{(-\infty,u]}(x)\phi(x)[/math]. The map [math]u\mapsto f(u,x)[/math] is continuous at [math]u_0[/math] for all [math]x\in\R\setminus\{u_0\}[/math] for some [math]u_0\in\R[/math]. But [math]\mu(\{u_0\})=0[/math], which implies that the map [math]u\mapsto f(u,x)[/math] is a.e. continuous at [math]u_0[/math] with [math]\vert f(u,x)\vert\leq \phi(x)\vert[/math] and [math]\vert\phi\vert[/math] is integrable by assumption.
• Consider now the Lebesgue measure [math]\lambda[/math] and [math]\phi\in\mathcal{L}^1(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)[/math]. Define moreover
  [[math]] \hat{\phi}(u)=\int_{\mathbb{R}}e^{iux}\phi(x)d\lambda, [[/math]]
  which is called the Fourier-transform of [math]\phi[/math]. The map [math]u\mapsto e^{iux}\phi(x)[/math] is actually continuous for all [math]x\in \R[/math], which implies that [math]\hat{\phi}(u)[/math] is continuous at any [math]u\in\mathbb{R}[/math].
• Let again [math]\phi\in\mathcal{L}^1(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)[/math] and [math]h:\mathbb{R}\longrightarrow \mathbb{R}[/math] be a continuous and bounded map. The convolution of [math]h[/math] and [math]\phi[/math] is given by
  [[math]] (h*\phi)(u)=\int_{\mathbb{R}}h(u-x)\phi(x)d\lambda(x). [[/math]]
  The map [math](h*\phi)[/math] is continuous at any [math]u\in\mathbb{R}[/math]. Moreover, the map [math]u\mapsto h(u-x)\phi(x)[/math] is continuous for all [math]x\in\mathbb{R}[/math]. Since [math]h[/math] is bounded, there exists a constant [math]k \gt 0[/math], such that [math]\vert h(x)\vert\leq k[/math] for all [math]x\in\R[/math].

Differentiation of Parameter Integrals

Theorem (Differentiation of Integrals)

Let [math]I\in \R[/math] be a real interval and [math](E,\A,\mu)[/math] a measure space. Let [math]f:I\times E\longrightarrow \Big|^\R_\C[/math] and let [math]u_0\in I[/math]. Moreover, assume that the following hold.

• The map [math]x\mapsto f(u,x)[/math] is in [math]\mathcal{L}^1(E,\A,\mu)[/math] for all [math]u\in I[/math].
• The map [math]u\mapsto f(u,x)[/math] is a.e. differentiable at [math]u_0[/math] with derivation denoted by [math]\partial_uf(u_0,x)[/math].
• There exists a map [math]g\in\mathcal{L}^1(E,\A,\mu)[/math] such that for all [math]u\in I[/math]
  [[math]] \vert f(u,x)-f(u_0,x)\vert\leq g(x)\vert u-u_0\vert. [[/math]]

Then the map [math]F(u)=\int_Ef(u,x)d\mu[/math] is differentiable at [math]u_0[/math] and its derivative is given by

[[math]] F'(u)=\int_E\partial_uf(u_0,x)d\mu. [[/math]]

It is often useful to replace [math](ii)[/math] and [math](iii)[/math] with the following points respectively.

• The map [math]u\mapsto f(u,x)[/math] is a.e. differentiable at any point in [math]I[/math].
• There exists a map [math]g\in\mathcal{L}^1(E,\mathcal{A},\mu)[/math] such that [math]\vert\partial_uf(u,x)\vert\leq g(x)[/math] a.e. for all [math]u\in I[/math].

Moreover, if [math]f[/math] is differentiable on a interval [math][a,b][/math], there exists a [math]\theta_{a,b}[/math], by the mean value theorem, such that

[[math]] f'(\theta_{a,b})=\frac{f(b)-f(a)}{b-a}. [[/math]]

If [math]f'[/math] is also bounded on [math][a,b][/math], i.e. there is a [math]M[/math] such that [math]\vert f'(x)\vert\leq M[/math], then

[[math]] \left\vert\frac{f(b)-f(a)}{b-a}\right\vert\leq M. [[/math]]

Show Proof

Let [math](u_n)_{n\geq 1}[/math] be a sequence in [math]I[/math] such that [math]u_n\xrightarrow{n\to\infty}u_0[/math] and assume that [math]u_n\not=u_0[/math] for all [math]n\geq 1[/math]. Now define the sequence

[[math]] \phi_n(x).=\frac{f(u_n,x)-f(u_0,x)}{u_n-u_0}. [[/math]]

Then we can obtain that

[[math]] \lim_{n\to\infty}\phi_n(x)=\partial_uf(u_0,x)\text{a.e.} [[/math]]

Now [math](iii)[/math] allows us to use Lebesgue's dominated convergence theorem. Hence we get

[[math]] \lim_{n\to\infty}\frac{F(u_n)-F(u_0)}{u_n-u_0}=\lim_{n\to\infty}\int_E\phi_n(x)d\mu=\int_E\partial_uf(u_0,x)d\mu. [[/math]]

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Example

[Fourier analysis] Let us give the following examples of Fourier analysis.

• Let [math]\phi\in \mathcal{L}^1(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)[/math] be an integrable map such that
  [[math]] \int_\mathbb{R}\vert x\phi(x)\vert d\lambda \lt \infty. [[/math]]
  Then its Fourier-transform is given by
  [[math]] \hat{\phi}(u)=\int_\mathbb{R}\underbrace{e^{iux}\phi(x)}_{f(u,x)}d\lambda, [[/math]]
  which is differentiable and the derivative of it is then
  [[math]] \hat{\phi}'(u)=i\int_{\mathbb{R}}e^{iux}x\phi(x)d\lambda. [[/math]]
  Therefore we can write
  [[math]] \vert\partial_uf(u,x)\vert=\vert ie^{iux}x\phi(x)\vert=\vert x\phi(x)\vert. [[/math]]
• Let [math]\phi\in \mathcal{L}^1(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)[/math] and [math]h:\mathbb{R}\longrightarrow \mathbb{R}[/math] be a bounded [math]C^1[/math]-map with bounded derivative [math]h'[/math]. Then the convolution [math](h*\phi)[/math] is differentiable and its derivative is given by [math](h*\phi)'=h'*\phi'[/math]. Recall that the convolution [math](h*\phi)[/math] is given by
  [[math]] (h*\phi)(u)=\int_{\mathbb{R}}\underbrace{h(u-x)\phi(x)}_{f(u,x)}d\lambda. [[/math]]
  Moreover, [math]\vert h(u-x)\phi(x)\vert\leq M\vert\phi(x)\vert[/math], where [math]M[/math] is such that [math]\vert h(x)\vert\leq M[/math] and [math]\vert h'(x)\vert\leq M[/math] for all [math]x\in\R[/math]. Then
  [[math]] \partial_uf(u,x)=h'(u-x)\phi(x),\text{and}\vert\partial_uf(u,x)\vert \leq M\vert\phi(x)\vert. [[/math]]

General references

Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].