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Measure theory and the notion of integration require special structures on functions, which need to satisfy different properties, such as being measurable or bounded. The notion of a measurable map is important for the study of the integration with respect to a certain measure and the fact that we can only consider integration with respect to a measure if the integrating function satisfies measurability. It is now important to use the <math>\sigma</math>-Algebras of the underlying spaces similar to the topological notion of continuity where the topologies of the underlying spaces are used. Let us therefore define a measurable map. | |||
{{definitioncard|Measurable Map| | |||
Let <math>(E,\A)</math> and <math>(F,\B)</math> be two measurable spaces and let <math>f:E\longrightarrow F</math> be a map. We say that <math>f</math> is measurable, if for all <math>B\in\B</math> we get <math>f^{-1}(B)\in\A</math>. | |||
}} | |||
{{proofcard|Proposition|prop-1|Let <math>(X,\A),(Y,\B)</math> and <math>(Z,\mathcal{G})</math> be measurable space and consider the composition | |||
<math display="block"> | |||
(X,\A)\xrightarrow{f}(X,\B)\xrightarrow{g}(Z,\mathcal{G}). | |||
</math> | |||
If <math>f</math> and <math>g</math> are both measurable, then <math>g\circ f</math> is also measurable. | |||
|Exercise.{{efn|Use the definition of a measurable map.}}}} | |||
{{proofcard|Proposition|prop-2|Let <math>(E,\A)</math> and <math>(F,\B)</math> be two measurable spaces and let <math>f:E\longrightarrow F</math> be a map. Moreover, assume that there exists <math>\mathcal{C}\subset\mathcal{P}(F)</math> such that <math>\sigma(\mathcal{C})=\B</math>. Then <math>f</math> is measurable if and only if for all <math>C\in\mathcal{C}</math> we have <math>f^{-1}(C)\in\A</math>. | |||
|Let us first define the <math>\sigma</math>-Algebra <math>\mathcal{G}</math> by | |||
<math display="block"> | |||
\mathcal{G}:=\{B\in\B\mid f^{-1}(B)\in\A\}\supset \mathcal{C}. | |||
</math> | |||
Now, since <math>\mathcal{G}</math> is a <math>\sigma</math>-Algebra, we get that <math>\sigma(\mathcal{C})\subset\mathcal{G}</math> and thus <math>\mathcal{G}=\B</math>, which proves the claim.}} | |||
'''Example''' | |||
Let{{efn|We can also take <math>(\bar \R,\B(\bar \R))</math>.}} <math>(F,\B)=(\R,\B(\R))</math>. To show that <math>f</math> is measurable, it is enough to show either that <math>f^{-1}((a,b))\in\A</math> or <math>f^{-1}((-\infty,a))\in\A</math>, for <math>a,b\in\R</math> with <math>a < b</math>. | |||
'''Example''' | |||
[continuous maps are measurable] | |||
Assume that <math>E</math> and <math>F</math> are two metric spaces (or topological spaces), endowed with their Borel <math>\sigma</math>-Algebra respectively. Then <math>f</math> is measurable if for every open set <math>O</math> of <math>F</math> we have <math>f^{-1}(O)\in\B(E)</math>. In particular we can say that ''continuous maps are measurable maps''. | |||
'''Example''' | |||
Let <math>A\subset E</math> be a subset of <math>E</math>. Then the map <math>\one_A</math> is measurable if and only if <math>A\in\A</math>. | |||
{{alert-info | | |||
The notion of measurability of a map <math>f:E\longrightarrow F</math>, between two measurable spaces <math>(E,\A)</math> and <math>(F,\B)</math>, means that <math>f^{-1}(\B)\in \A</math>. The smallest <math>\sigma</math>-Algebra on <math>E</math> which makes <math>f</math> measurable is given by <math>f^{-1}(\B)</math> and we denote it by <math>\sigma(f)</math>. Moreover, we want to emphasize that we can write <math>\{f\in B\}</math> for <math>f^{-1}(B)=\{x\in E\mid f(x)\in B\}</math>. Hence we can write <math>\{f\geq b\}</math> instead of <math>f^{-1}([b,\infty))</math> or <math>\{ f=b\}</math> instead of <math>f^{-1}(\{b\})</math>. If <math>f</math> is constant and if for some <math>C\in F</math>, we have <math>f(x)=C</math> for all <math>x\in E</math>, then <math>f</math> is always measurable, since <math>f^{-1}(\B)=\{\varnothing, E\}</math>. | |||
}} | |||
{{proofcard|Lemma|lem-1|Let <math>(E,\A),(F_1,\B_1)</math> and <math>(F_2,\B_2)</math> be measurable spaces and let <math>f_1:E\longrightarrow F_1</math> and <math>f_2:E\longrightarrow F_2</math> be two measurable maps. Then the map | |||
<math display="block"> | |||
\begin{align*} | |||
f:(E,\A)&\longrightarrow (F_1\times F_2,\B_1\otimes \B_2)\\ | |||
x&\longmapsto (f_1(x),f_2(x)) | |||
\end{align*} | |||
</math> | |||
is measurable. | |||
|Let us define <math>\mathcal{C}:=\{B_1\times B_2\midB_1\in\mathcal{B}_1,B_2\in\mathcal{B}_2\}</math>. Then we get, by definition of the product <math>\sigma</math>-Algebra, that <math>\sigma(\mathcal{C})=\mathcal{B}_1\otimes\mathcal{B}_2</math>. Now, for <math>B_1\times B_2\in \mathcal{C}</math> we get that <math>f^{-1}(B_1\times B_2)=\underbrace{\underbrace{f^{-1}(B_1)}_{\in\mathcal{A}}\cap \underbrace{f^{-1}(B_2)}_{\in\mathcal{A}}}_{\in\mathcal{A}}\in\A</math>. Therefore, it follows that <math>f</math> is measurable.}} | |||
{{alert-info | | |||
Consider <math>f_1=\pi_1\circ f</math> and <math>f_2=\pi_2\circ f</math> with | |||
<math display="block"> | |||
\begin{align*} | |||
\pi_i:F_1\times F_2&\longrightarrow F_i\\ | |||
(y_1,y_2)&\longmapsto y_i | |||
\end{align*} | |||
</math> | |||
for <math>i\in\{1,2\}</math> with <math>\pi_1</math> and <math>\pi_2</math> are measurable. Then <math>f_1</math> and <math>f_2</math> are measurable. | |||
}} | |||
{{proofcard|Corollary|cor-1|Let <math>(E,\A)</math> be a measurable space and let <math>f,g:E\longrightarrow \R</math> be two measurable maps, where <math>\R</math> is endowed with its Borel <math>\sigma</math>-Algebra <math>\B(\R)</math>. Then | |||
<ul style{{=}}"list-style-type:lower-roman"><li><math>f+g</math> | |||
</li> | |||
<li><math>f\cdot g</math> | |||
</li> | |||
<li><math>f^+=\max\{f,0\}</math> | |||
</li> | |||
<li><math>f^-=\max\{-f,0\}</math> | |||
</li> | |||
<li><math>\vert f\vert</math> | |||
</li> | |||
</ul> | |||
are measurable, where <math>f = f^+-f^-</math> and <math>\vert f\vert =f^++f^-</math>. | |||
|We will only show <math>(i)</math> and leave the other points as an exercise for the reader. The map <math>f+g</math> is a composition of the map <math>h:x\longmapsto (f(x),g(x))</math> and <math>r:(a,b)\longmapsto a+b</math>. The map <math>h</math> is clearly measurable, since <math>f</math> and <math>g</math> are measurable and the map <math>r</math> is clearly continuous and thus measurable. As we have seen, the composition of two measurable maps is again measurable, which shows that <math>f+g</math> is measurable. The proof of the other points is similar.}} | |||
{{alert-info | | |||
Let us consider the field of complex numbers <math>\C</math> and make the identification <math>\C\simeq \R^2</math>. Then we can naturally make sense of the measurability of the map <math>f:E\longrightarrow \C</math>, where <math>\C</math> is endowed with its Borel <math>\sigma</math>-Algebra <math>\B(\C)</math>, by saying that <math>f</math> is measurable if and only if <math>Re(f)</math> and <math>Im(f)</math> are measurable. | |||
}} | |||
==General references== | |||
{{cite arXiv|last=Moshayedi|first=Nima|year=2020|title=Lectures on Probability Theory|eprint=2010.16280|class=math.PR}} | |||
==Notes== | |||
{{notelist}} |
Revision as of 01:53, 8 May 2024
Measure theory and the notion of integration require special structures on functions, which need to satisfy different properties, such as being measurable or bounded. The notion of a measurable map is important for the study of the integration with respect to a certain measure and the fact that we can only consider integration with respect to a measure if the integrating function satisfies measurability. It is now important to use the [math]\sigma[/math]-Algebras of the underlying spaces similar to the topological notion of continuity where the topologies of the underlying spaces are used. Let us therefore define a measurable map.
Let [math](E,\A)[/math] and [math](F,\B)[/math] be two measurable spaces and let [math]f:E\longrightarrow F[/math] be a map. We say that [math]f[/math] is measurable, if for all [math]B\in\B[/math] we get [math]f^{-1}(B)\in\A[/math].
Let [math](X,\A),(Y,\B)[/math] and [math](Z,\mathcal{G})[/math] be measurable space and consider the composition
Exercise.[a]
Let [math](E,\A)[/math] and [math](F,\B)[/math] be two measurable spaces and let [math]f:E\longrightarrow F[/math] be a map. Moreover, assume that there exists [math]\mathcal{C}\subset\mathcal{P}(F)[/math] such that [math]\sigma(\mathcal{C})=\B[/math]. Then [math]f[/math] is measurable if and only if for all [math]C\in\mathcal{C}[/math] we have [math]f^{-1}(C)\in\A[/math].
Let us first define the [math]\sigma[/math]-Algebra [math]\mathcal{G}[/math] by
Example
Let[b] [math](F,\B)=(\R,\B(\R))[/math]. To show that [math]f[/math] is measurable, it is enough to show either that [math]f^{-1}((a,b))\in\A[/math] or [math]f^{-1}((-\infty,a))\in\A[/math], for [math]a,b\in\R[/math] with [math]a \lt b[/math].
Example
[continuous maps are measurable] Assume that [math]E[/math] and [math]F[/math] are two metric spaces (or topological spaces), endowed with their Borel [math]\sigma[/math]-Algebra respectively. Then [math]f[/math] is measurable if for every open set [math]O[/math] of [math]F[/math] we have [math]f^{-1}(O)\in\B(E)[/math]. In particular we can say that continuous maps are measurable maps.
Example
Let [math]A\subset E[/math] be a subset of [math]E[/math]. Then the map [math]\one_A[/math] is measurable if and only if [math]A\in\A[/math].
The notion of measurability of a map [math]f:E\longrightarrow F[/math], between two measurable spaces [math](E,\A)[/math] and [math](F,\B)[/math], means that [math]f^{-1}(\B)\in \A[/math]. The smallest [math]\sigma[/math]-Algebra on [math]E[/math] which makes [math]f[/math] measurable is given by [math]f^{-1}(\B)[/math] and we denote it by [math]\sigma(f)[/math]. Moreover, we want to emphasize that we can write [math]\{f\in B\}[/math] for [math]f^{-1}(B)=\{x\in E\mid f(x)\in B\}[/math]. Hence we can write [math]\{f\geq b\}[/math] instead of [math]f^{-1}([b,\infty))[/math] or [math]\{ f=b\}[/math] instead of [math]f^{-1}(\{b\})[/math]. If [math]f[/math] is constant and if for some [math]C\in F[/math], we have [math]f(x)=C[/math] for all [math]x\in E[/math], then [math]f[/math] is always measurable, since [math]f^{-1}(\B)=\{\varnothing, E\}[/math].
Let [math](E,\A),(F_1,\B_1)[/math] and [math](F_2,\B_2)[/math] be measurable spaces and let [math]f_1:E\longrightarrow F_1[/math] and [math]f_2:E\longrightarrow F_2[/math] be two measurable maps. Then the map
is measurable.
Let us define [math]\mathcal{C}:=\{B_1\times B_2\midB_1\in\mathcal{B}_1,B_2\in\mathcal{B}_2\}[/math]. Then we get, by definition of the product [math]\sigma[/math]-Algebra, that [math]\sigma(\mathcal{C})=\mathcal{B}_1\otimes\mathcal{B}_2[/math]. Now, for [math]B_1\times B_2\in \mathcal{C}[/math] we get that [math]f^{-1}(B_1\times B_2)=\underbrace{\underbrace{f^{-1}(B_1)}_{\in\mathcal{A}}\cap \underbrace{f^{-1}(B_2)}_{\in\mathcal{A}}}_{\in\mathcal{A}}\in\A[/math]. Therefore, it follows that [math]f[/math] is measurable.
Consider [math]f_1=\pi_1\circ f[/math] and [math]f_2=\pi_2\circ f[/math] with
for [math]i\in\{1,2\}[/math] with [math]\pi_1[/math] and [math]\pi_2[/math] are measurable. Then [math]f_1[/math] and [math]f_2[/math] are measurable.
Let [math](E,\A)[/math] be a measurable space and let [math]f,g:E\longrightarrow \R[/math] be two measurable maps, where [math]\R[/math] is endowed with its Borel [math]\sigma[/math]-Algebra [math]\B(\R)[/math]. Then
- [math]f+g[/math]
- [math]f\cdot g[/math]
- [math]f^+=\max\{f,0\}[/math]
- [math]f^-=\max\{-f,0\}[/math]
- [math]\vert f\vert[/math]
are measurable, where [math]f = f^+-f^-[/math] and [math]\vert f\vert =f^++f^-[/math].
We will only show [math](i)[/math] and leave the other points as an exercise for the reader. The map [math]f+g[/math] is a composition of the map [math]h:x\longmapsto (f(x),g(x))[/math] and [math]r:(a,b)\longmapsto a+b[/math]. The map [math]h[/math] is clearly measurable, since [math]f[/math] and [math]g[/math] are measurable and the map [math]r[/math] is clearly continuous and thus measurable. As we have seen, the composition of two measurable maps is again measurable, which shows that [math]f+g[/math] is measurable. The proof of the other points is similar.
Let us consider the field of complex numbers [math]\C[/math] and make the identification [math]\C\simeq \R^2[/math]. Then we can naturally make sense of the measurability of the map [math]f:E\longrightarrow \C[/math], where [math]\C[/math] is endowed with its Borel [math]\sigma[/math]-Algebra [math]\B(\C)[/math], by saying that [math]f[/math] is measurable if and only if [math]Re(f)[/math] and [math]Im(f)[/math] are measurable.
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].