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

= 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 Tn is measurable? &mdash; 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
 * $$W_\alpha = \bigcup_k V_k = \{x\ |\ \sum_{k=1}^\infty f_k(x) > \alpha\}$$

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 $$\alpha = 1/n$$. &mdash; merge 22:30, 16 March 2008 (UTC)