 Nicholas Cruz

2022-05-23

MathJax(?): Can't find handler for document 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 $a:\mathrm{\Omega }\to {\mathbb{R}}_{+}\cup \left\{\mathrm{\infty }\right\}$ be a map. Then we define
$\sum _{\mathrm{\Omega }}a\left(\omega \right):=\underset{F\subset \mathrm{\Omega },|F|<\mathrm{\infty }}{sup}\sum _{F}a\left(\omega \right)$

Now our Prof said that we can consider $\sum _{\mathrm{\Omega }}a\left(\omega \right)$ 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 $f:\mathrm{\Omega }\to {\mathbb{R}}_{+}\cup \left\{\mathrm{\infty }\right\}$ where $f\left(\mathrm{\Omega }\right)=\left\{{b}_{1},...,{b}_{n}\right\}$ finately many then

where ${\mathrm{\Omega }}_{i}={f}^{-1}\left(\left\{{b}_{i}\right\}\right)$. 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 $\mathrm{\Omega }$. Sasha Pacheco

Expert

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