Testwiki: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

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

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

$T_n=\{x|1/n\le \sum _k\nu (R_{k,x})\}.$

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

$W_{\alpha }=\bigcup _k{V}_k=\{x|\sum _{k=1}^{\mathrm{\infty }}{f}_k(x)>\alpha \}$

and

$\bigcap _j{W}_{\alpha -1/j}=\{x|\sum _{k=1}^{\mathrm{\infty }}{f}_k(x)\ge \alpha \}$.

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)