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

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March 15[edit]

A measurable set?[edit]

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

Well, R_{k,x} is just Bk if x is in Ak, and empty otherwise. So \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 \{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\}


\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)