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

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March 15

A measurable set?

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

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

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

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

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

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

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

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

are measurable, and so are

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

and

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

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