March 15
Let
and
be measure spaces and
sequences of sets of finite measure in X and Y respectively. Let the "rectangles"
, and assume that
![{\displaystyle \sum _{k=1}^{\infty }(\mu \otimes \nu )(R_{k})<\infty .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c19f49971203088cf5e4896934bc955704ba26c)
Let
and
![{\displaystyle T_{n}=\{x\ |\ 1/n\leq \sum _{k}\nu (R_{k,x})\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56f1ebdbf04e3a1a5bbf1cb05dab1f10983625ed)
Why is it obvious that Tn is measurable? — merge 17:01, 15 March 2008 (UTC)[reply]
- Well,
is just Bk if x is in Ak, and empty otherwise. So
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)[reply]
Oh, I think I see how it works out. If
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 \))](https://wikimedia.org/api/rest_v1/media/math/render/svg/26019a4981cc2c96d45989f6652ea2af4d379c10)
are measurable, and so are
![{\displaystyle W_{\alpha }=\bigcup _{k}V_{k}=\{x\ |\ \sum _{k=1}^{\infty }f_{k}(x)>\alpha \))](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d9ba43aea8022933f719dae4db91da1ffa6a5c1)
and
.
In this case
and
. — merge 22:30, 16 March 2008 (UTC)[reply]