# Wikipedia:Reference desk/Archives/Mathematics/2008 March 15

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# March 15

## A measurable set?

Let ${\displaystyle (X,\mu )}$ and ${\displaystyle (Y,\nu )}$ be measure spaces and ${\displaystyle \{A_{k}\},\{B_{k}\}}$ sequences of sets of finite measure in X and Y respectively. Let the "rectangles" ${\displaystyle R_{k}=A_{k}\times B_{k}}$, and assume that

${\displaystyle \sum _{k=1}^{\infty }(\mu \otimes \nu )(R_{k})<\infty .}$

Let ${\displaystyle R_{k,x}=\{y\ |\ (x,y)\in R_{k}\}}$ and

${\displaystyle T_{n}=\{x\ |\ 1/n\leq \sum _{k}\nu (R_{k,x})\}.}$

Why is it obvious that Tn is measurable?  — merge 17:01, 15 March 2008 (UTC)

Well, ${\displaystyle R_{k,x}}$ is just Bk if x is in Ak, and empty otherwise. So ${\displaystyle \sum _{k}\nu (R_{k,x})}$ is the some of the measures of the Bk such that x is in Ak. Thus whether x is in Tn is determined by which of the Aks x is in, and Tn is a union of intersections of the Aks. Algebraist 17:45, 15 March 2008 (UTC)

Oh, I think I see how it works out. If ${\displaystyle \{f_{k}\}}$ is a sequence of nonnegative measurable real-valued functions and α is a real number, the sets

${\displaystyle V_{k}=\{x\ |\ f_{1}(x)+\cdots +f_{k}(x)>\alpha \}}$

are measurable, and so are

${\displaystyle W_{\alpha }=\bigcup _{k}V_{k}=\{x\ |\ \sum _{k=1}^{\infty }f_{k}(x)>\alpha \}}$

and

${\displaystyle \bigcap _{j}W_{\alpha -1/j}=\{x\ |\ \sum _{k=1}^{\infty }f_{k}(x)\geq \alpha \}}$.

In this case ${\displaystyle f_{k}(x)=\nu (R_{k,x})=\chi _{A_{k}}(x)\nu (B_{k})}$ and ${\displaystyle \alpha =1/n}$.  — merge 22:30, 16 March 2008 (UTC)