Integration with respect to a positive measure
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
There is no way where we can say that this function would be Riemann integrable and thus we cant make sense of the integral
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
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.
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
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
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
On the other hand, we can also write [math]f[/math] as
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
and therefore the integral doesn't depend on the representation of [math]f[/math].
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]]
For [math](i)[/math], let us first set
Moreover, we can note that
and therefore we can write
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
It follows that
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
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
Sometimes we have different notation for the same.
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]]
Exercise.[a]
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
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
We need to show that
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
We see immediately that [math]E_n[/math] is measurable[b] and since [math]f_n\uparrow f[/math] as [math]n\to\infty[/math] and [math]a \lt 1[/math], we see that
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
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
Note that the left side doesn't depend on [math]a[/math]. Now we let [math]a\to 1[/math] and obtain then
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
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].
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]]
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
where [math]MCV[/math] stands for monotone convergence. Now [math](ii)[/math] is an Immediate consequence of the monotone convergence theorem (MCV). Indeed, set
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
which proves the proposition.
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
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
Moreover, we also get for every positive sequence [math](a_{n,k})_{n\in\mathbb{N},k\in\mathbb{N}}\geq 0[/math] that
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
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
It is clear that if [math]\mu(A)=0[/math] then [math]\nu(A)=0[/math] for all [math]A\in\A[/math].
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
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
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].
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]]
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
Recall that we actually have
From the monotone convergence theorem (MCV) we get
Now for all [math]p\geq k[/math] we have
Integrable functions
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
Moreover, if [math]f[/math] is integrable, we define its integral to be given by
For instance
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.
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]
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].
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
and we write [math]f\in\mathcal{L}^1_\C(E,\A,\mu)[/math]. This is simply because of the fact that
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.
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
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
Then [math]\mu(A^C)=0[/math]. Let us now apply the first part of the proof to
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
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
for some [math]u\in E[/math], which actually gives rise to a map
Example
[Gamma function] Let us start with a special example of such a map. The Gamma function is defined as the map
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
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].
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].
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
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
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
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
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
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].