Simple Functions

[math] \newcommand{\R}{\mathbb{R}} \newcommand{\A}{\mathcal{A}} \newcommand{\B}{\mathcal{B}} \newcommand{\N}{\mathbb{N}} \newcommand{\C}{\mathbb{C}} \newcommand{\Rbar}{\overline{\mathbb{R}}} \newcommand{\Bbar}{\overline{\mathcal{B}}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\E}{\mathbb{E}} \newcommand{\p}{\mathbb{P}} \newcommand{\one}{\mathds{1}} \newcommand{\0}{\mathcal{O}} \newcommand{\mat}{\textnormal{Mat}} \newcommand{\sign}{\textnormal{sign}} \newcommand{\CP}{\mathcal{P}} \newcommand{\CT}{\mathcal{T}} \newcommand{\CY}{\mathcal{Y}} \newcommand{\F}{\mathcal{F}} \newcommand{\mathds}{\mathbb}[/math]

After we have developed the notion of a measurable map, we need to discuss a class of very powerful and, as the name points out, simple functions. The advantage of these type of functions are exactly the fact that they are simple to handle and moreover one can basically proof many things for measurable functions by proving it for simple functions and deduce the general case out of that. We will later see the advantage of them being dense in different spaces. Let us start with the definition of a simple function.

Definition (Simple function)

Let [math](E,\A)[/math] be a measurable space. A map [math]f:E\longrightarrow \R[/math] is called simple, if it is measurable and if it takes a finite number of values. Recall again that [math]\R[/math] is considered as a measurable space endowed with its Borel [math]\sigma[/math]-Algebra [math]\B(\R)[/math].

Remark

By definition, one can therefore write any simple function [math]f[/math] as

[[math]] f=\sum_{i\in I}\alpha_i\one_{A_i}, [[/math]]

where [math]I[/math] is a finite index set, [math]\alpha_i[/math] are real numbers and the sets [math](A_i)_{i\in I}[/math] form a [math]\A[/math]-measurable partition of [math]E[/math], i.e. [math]E=\bigcup_{i\in I}A_i[/math], [math]A_i\cap A_j=\varnothing[/math] if [math]i\not=j[/math] and [math]A_i\in \A[/math] for all [math]i\in I[/math].

Show Proof

Note first that for all [math]x\in E[/math], we get that [math]f(x)\in\{\alpha_1,...,\alpha_k\}[/math], where the [math]\alpha_i[/math]'s are distinct if and only if [math]f^{-1}(\{\alpha_i\})=\{x\in E\mid f(x)=\alpha_i\}=A_i\in\A[/math], since [math]f[/math] is measurable. Hence we get that [math]A_i\cap A_j=\varnothing[/math] for [math]i\not=j[/math] and [math]\bigcup_{i=1}^kA_i=E[/math]. This representations is unique if the [math]\alpha_i's[/math] are distinct. We can write the canonical form therefore as

[[math]] f=\sum_{\alpha\in f(E)}\alpha\one_{f=\alpha}. [[/math]]

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We can notice the fact that the simple functions form a commutative algebra. indeed, let [math]f=\sum_{i\in I}\alpha_i\one_{A_i}[/math] and [math]g=\sum_{j\in J}\beta_j\one_{B_j}[/math] be two simple functions (in canonical form) and let [math]\lambda\in\R[/math]. Then we can easily obtain that

[[math]] \lambda f+g=\sum_{i\in I\atop j\in J}(\lambda\alpha_i+\beta_j)\one_{A_i\cap B_j} [[/math]]

and since [math](A_i\cap B_j)_{(i,j)\in I\times J}[/math] forms an [math]\A[/math]-partition of [math]E[/math], we get that

[[math]] f\cdot g=\sum_{i\in I\atop j\in J}\alpha_i\beta_j\one_{A_i\cap B_j} [[/math]]

is a simple function and moreover,

[[math]] \max(f,g)=\sum_{i\in I\atop j\in J}\max(\alpha_i,\beta_j)\one_{A_i\cap B_j}. [[/math]]

Theorem

Let [math](E,\A)[/math] be a measurable space and let [math]f:E\longrightarrow \R[/math] be a measurable map. Then there exists a sequence of simple functions [math](f_n)_{n\in\N}[/math] such that for all [math]x\in E[/math] we get

[[math]] f_n(x)\xrightarrow{n\to\infty}f(x). [[/math]]

Moreover, if

• [math]f\geq 0[/math], we can choose an increasing [math]f_n\geq 1[/math] ([math]0\leq f_n\leq f_{n+1}[/math]).
• [math]f[/math] is bounded, [math]f_n[/math] can be chosen such that the convergence is uniformly, i.e.
  [[math]] \sup_{x\in E}\vert f_n(x)-f(x)\vert\xrightarrow{n\to\infty}0. [[/math]]

Show Proof

Let us first assume that [math]f\geq 0[/math]. For [math]n\in\mathbb{N}[/math], we set

[[math]] E_{n,\infty}:=\{f\geq n\},E_{n,k}:=\left\{\frac{k}{2^n}\leq f \lt \frac{k+1}{2^n}\right\},k\in\{0,1,...,n2^n-1\}. [[/math]]

Thus we have [math]E_{n,k},E_{n,\infty}\in\mathcal{A}[/math]. Now define

[[math]] f_n:=\sum_{k=0}^{n2^n-1}\frac{k}{2^n}\cdot\one_{E_{n,k}}+n\one_{E_{n,\infty}} [[/math]]

and obtain that [math]f_n[/math] is simple by construction. For [math]x\in E_{n,k}[/math] we get

[[math]] f_{n+1}(x)=\begin{cases}f_n(x), &\frac{2k}{2^{n+1}}\leq f(x) \lt \frac{2k+1}{2^{n+1}}\\ f_n(x)+\frac{1}{2^{n+1}}, &\frac{2k+1}{2^{n+1}}\leq f(x) \lt \frac{2(k+1)}{2^{n+1}}\end{cases} [[/math]]

and if [math]x\in E_{n,\infty}[/math] we get

[[math]] f_{n+1}(x)=\begin{cases} n+1, & f(x)\geq n+1\\ \frac{n2^{n+1}+l}{2^{n+1}}, &\frac{n2^{n+1}+l}{2^{n+1}}\leq f(x) \lt \frac{n2^{n+1}+l+1}{2^{n+1}}\end{cases} [[/math]]

It follows that [math]0\leq f_n(x)\leq f_{n+1}(x)[/math] for all [math] x\in E[/math]. If furthermore [math]x\in \{f \lt n\}[/math], then

[[math]] 0\leq f(x)-f_n(x)\leq 2^{-n}=\frac{1}{2^n}=\frac{k+1}{2^n}-\frac{k}{2^n}\xrightarrow{n\to\infty} 0 [[/math]]

or equivalently

[[math]] f_n(x)\xrightarrow{n\to\infty} f(x) [[/math]]

on [math]\left\{x\in E\mid f(x) \lt \infty\right\}=\bigcup_{k\geq 1}\{f \lt k\}[/math], and if [math]x\in \{f=\infty\}=\bigcap_{n\in\mathbb{N}}\{f\geq n\}[/math], then

[[math]] f_n(x)=n\xrightarrow{n\to\infty}\infty. [[/math]]

If we have a function [math]f:E\longrightarrow \mathbb{R}^+[/math] and we assume that there exists aome [math]M\in(0,\infty)[/math] such that [math]0\leq f_n(x)\leq M[/math] for all [math]x\in E[/math], then for [math]n \gt M[/math] with [math]\{f\geq n\}=\varnothing[/math] it follows that for all [math]x\in E[/math] we get

[[math]] 0\leq f(x)-f_n(x)\leq 2^{-n}, [[/math]]

which implies that

[[math]] \sup_{x\in E}\vert f(x)-f_n(x)\vert \leq 2^{-n}\xrightarrow{n\to\infty} 0. [[/math]]

Let us emphasize the real case. If we have a function [math]f:E\longrightarrow \overline{\mathbb{R}}[/math] we have the decompositions as

[[math]] f=f^+-f^-,\vert f\vert =f^1+f^-,\text{with}f^+,f^-\geq 0. [[/math]]

If we now take [math]f_n^+[/math] and [math]f_n^-[/math] as constructed above we can obtain [math]f_n^+ \uparrow f^+[/math] and [math]f_n^- \uparrow f^-[/math] for [math]n\to\infty[/math]. Moreover, we notice that for [math]x\in E[/math] the sequences [math](f_n^+(x))_{n\in\N}[/math] and [math](f_n^-(x))_{n\in\N}[/math] cannot be simultaneously nonzero and therefore

[[math]] f_n=f_n^+-f_n^-\xrightarrow{n\to\infty} f=f^+-f^-. [[/math]]

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General references

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