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How do I prove that a set is an element of a product $\sigma$-algebra?
We have $(X, m, \mu)$ and $([0,1), \operatorname{Bor}([0,1)), \lambda)$ are measure spaces ($\lambda$ is the Lebesgue measure) and (X,m,$\mu$) is a finite measure space. Also, $f$: X$\rightarrow$$[0,1$$)$ measurable.

$$U = \{(x, t) \in X \times [0,1) : 0 \le t < f(x)\}$$

How do I prove that $U\in m\otimes\operatorname{Bor}([0,1))$?

I've read that alternatively I could show that $X\times[0,1)$ is measurable. Why is this equivalent?

In order to show $X\times[0,1)$ is measurable do I show
$\forall x \in X$, $U_x=\{t\in [0,1) : (x,t) \in U \} \in \operatorname{Bor}([0,1))$
$\forall y \in [0,1)$, $U^y =\{x \in X : (x, t) \in U \} \in m$

Thanks for your help. Sorry about the formatting, I'm a newbie.

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I've begun moving this closer to standard TeX usage, but there's more to be done. –  Michael Hardy May 8 '13 at 21:51
What do you mean by measurable? With respect to what? It is the entire space, so it will always be measurable in the ways I can think of at the moment. –  tomasz May 8 '13 at 22:11
Also did you mean that $0\leq t<f(x)$ in the definition of $U$? –  tomasz May 8 '13 at 22:13
From the latter part of your post, it looks like you're trying to show that $U$ is measurable. In any case, it is not enough for all sections to be measurable for a set to be measurable. Even if all of them are singletons it may still not be the case. –  tomasz May 8 '13 at 22:16
If a set is measurable in the product sigma algebra, then its sections are measurable. This is an essential ingredient towards proving Fubini, for e.g.. Tomasz is saying that the converse is not true, necessarily. –  A Blumenthal May 8 '13 at 22:24

1 Answer 1

up vote 2 down vote accepted

Here's one method. You can show that $$U = \{(x,t) \mid 0 \leq t < f(x)\}$$ by forming $U$ as a countable union of elements from $m \otimes \text{Bor}[0,1)$. Recall that $f$ is the increasing limit of simple functions $f_n : X \rightarrow [0,1)$ of the form $$ f_n(x) = \sum_{k = 0}^{2^n} k 2^{-n} \cdot \chi_{f \in [2^{-n}k, 2^{-n}(k+1))}(x) $$ Let $$U_n = \{(x,t) \mid 0 \leq t < f_n(x)\} = \bigcup_{k = 0}^{2^n} \{f \in [2^{-n}k, 2^{-n}(k+1))\} \times [0,k 2^{-n})$$

Now you figure out the following:

  1. Are the $U_n$'s measurable in $m \otimes \text{Bor}[0,1)$?
  2. What is the relationship between the $U_n$'s and $U$?
  3. How does this relationship, combined with the axioms for a sigma algebra, prove that $U$ belongs to $m \otimes \text{Bor}[0,1)$?
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Ok, so intuitively I can see that the $U$_n's are measurable. Do I need ($U$$_n$)$_x$ and ($U$$_n$)$^y$ to prove that? and $U$ should be the union of the $U$$_n$'s. And I'm assuming that $U$ belongs to $m$ $\otimes$ $Bor$[0,1) since for all n $U$$_n$ $\epsilon$ $m$ $\otimes$ $Bor$[0,1). But I guess my main question is how do we know that? There seems to be a link between being measurable and being an element that I am missing. Thanks! –  Catfish May 8 '13 at 23:22
@Catfish The key is how $m \otimes \text{Bor}[0,1)$ is defined: it's the smallest sigma algebra over $X \times [0,1)$ for which all measurable rectangles $A \times B$ are measurable, when $A \in m, B \in \text{Bor}[0,1)$. In particular, finite and countable unions of rectangles are measurable in $m \otimes \text{Bor}[0,1)$. –  A Blumenthal May 9 '13 at 0:51
The moral of the comments under your question is that the measurability of sections tells you nothing about measurability in $m \otimes \text{Bor}[0,1)$. –  A Blumenthal May 9 '13 at 0:52
@A Blumenthal That was exactly what I was missing! Thanks so much! –  Catfish May 9 '13 at 12:43

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