Nicholas Cruz

Answered

2022-05-23

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MathJax(?): Can't find handler for document
I participate the stochastic course and we now speak about summable families. There we have the following definition:

Let $\mathrm{\Omega}$ be countable and $a:\mathrm{\Omega}\to {\mathbb{R}}_{+}\cup \{\mathrm{\infty}\}$ be a map. Then we define

$\sum _{\mathrm{\Omega}}a(\omega ):=\underset{F\subset \mathrm{\Omega},|F|<\mathrm{\infty}}{sup}\sum _{F}a(\omega )$

Now our Prof said that we can consider $\sum _{\mathrm{\Omega}}a(\omega )$ as the integral of math xmlns="http://www.w3.org/1998/Math/MathML">a over $\mathrm{\Omega}$ to get a better connetion to measure theory afterwards when we speak about Fatou's lemma, Beppo Levi theorem ect. Because all this theorems we have seen with integrals last semester.

But somehow I don't see why this sum is equal to the integral. So I know from measure theory that if we have a simple function $f:\mathrm{\Omega}\to {\mathbb{R}}_{+}\cup \{\mathrm{\infty}\}$ where $f(\mathrm{\Omega})=\{{b}_{1},...,{b}_{n}\}$ finately many then

${\int}_{\mathrm{\Omega}}f\text{}\mathsf{d}\mu =\sum _{i=1}^{n}f({b}_{i})\cdot \mu ({\mathrm{\Omega}}_{i})$

where ${\mathrm{\Omega}}_{i}={f}^{-1}(\{{b}_{i}\})$. So but here I don't think that this has to do something with simple functions right?

Therefore I wanted to ask you if someone could explain me why we can see this sum as the integral of $a$ over $\mathrm{\Omega}$.

Let $\mathrm{\Omega}$ be countable and $a:\mathrm{\Omega}\to {\mathbb{R}}_{+}\cup \{\mathrm{\infty}\}$ be a map. Then we define

$\sum _{\mathrm{\Omega}}a(\omega ):=\underset{F\subset \mathrm{\Omega},|F|<\mathrm{\infty}}{sup}\sum _{F}a(\omega )$

Now our Prof said that we can consider $\sum _{\mathrm{\Omega}}a(\omega )$ as the integral of math xmlns="http://www.w3.org/1998/Math/MathML">

But somehow I don't see why this sum is equal to the integral. So I know from measure theory that if we have a simple function $f:\mathrm{\Omega}\to {\mathbb{R}}_{+}\cup \{\mathrm{\infty}\}$ where $f(\mathrm{\Omega})=\{{b}_{1},...,{b}_{n}\}$ finately many then

${\int}_{\mathrm{\Omega}}f\text{}\mathsf{d}\mu =\sum _{i=1}^{n}f({b}_{i})\cdot \mu ({\mathrm{\Omega}}_{i})$

where ${\mathrm{\Omega}}_{i}={f}^{-1}(\{{b}_{i}\})$. So but here I don't think that this has to do something with simple functions right?

Therefore I wanted to ask you if someone could explain me why we can see this sum as the integral of $a$ over $\mathrm{\Omega}$.

Answer & Explanation

Sasha Pacheco

Expert

2022-05-24Added 10 answers

First prove that if $\mathrm{\Omega}=\{{\omega}_{1},{\omega}_{2},\dots \}$, then $\sum _{\omega \in \mathrm{\Omega}}a(\omega )=\sum _{i=1}^{\mathrm{\infty}}a({\omega}_{i})$.

Now take $\mu $ to be the counting measure on $\mathrm{\Omega}$. Really, without loss of generality, you can assume $\mathrm{\Omega}=\mathbb{N}$, so you want to prove that ${\int}_{\mathbb{N}}a\phantom{\rule{thinmathspace}{0ex}}d\mu =\sum _{i=1}^{\mathrm{\infty}}{a}_{i}$. This is obvious if there exists $N$ such that ${a}_{i}=0$ for $i\ge N$. The general case follows from monotone convergence theorem.

Now take $\mu $ to be the counting measure on $\mathrm{\Omega}$. Really, without loss of generality, you can assume $\mathrm{\Omega}=\mathbb{N}$, so you want to prove that ${\int}_{\mathbb{N}}a\phantom{\rule{thinmathspace}{0ex}}d\mu =\sum _{i=1}^{\mathrm{\infty}}{a}_{i}$. This is obvious if there exists $N$ such that ${a}_{i}=0$ for $i\ge N$. The general case follows from monotone convergence theorem.

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