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Nicholas Cruz

Nicholas Cruz

Answered

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 Ω be countable and a : Ω R + { } be a map. Then we define
Ω a ( ω ) := sup F Ω , | F | < F a ( ω )

Now our Prof said that we can consider Ω a ( ω ) as the integral of math xmlns="http://www.w3.org/1998/Math/MathML"> a over Ω 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 : Ω R + { } where f ( Ω ) = { b 1 , . . . , b n } finately many then
Ω f   d μ = i = 1 n f ( b i ) μ ( Ω i )
where Ω 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 Ω.

Answer & Explanation

Sasha Pacheco

Sasha Pacheco

Expert

2022-05-24Added 10 answers

First prove that if Ω = { ω 1 , ω 2 , }, then ω Ω a ( ω ) = i = 1 a ( ω i ).
Now take μ to be the counting measure on Ω. Really, without loss of generality, you can assume Ω = N , so you want to prove that N a d μ = i = 1 a i . This is obvious if there exists N such that a i = 0 for i N. The general case follows from monotone convergence theorem.

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