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If $h(x,t)$ is measurable and the measures involved are $\sigma$-finite, does there exist a sequences of functions

$$h_n(x,t) = \sum_{j=1}^{N_n} f_{j,n}(x) 1_{F_{j,n}}(t)$$ where the sets are pairwise disjoint (in $j$) such that $h_n \to h$ pointwise almost everywhere?

I know that if we fix $x$ we can get a sequence like that but we cannot just make the coefficients depending on $x$ since the sets $F_{j,n}$ will be different. Can I somehow combine them? Do I need the $\sigma$-finiteness?

If the answer is positive, we can use this to prove Minkowski's inequality for integrals quite easily.

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Sorry if it is obvious to those who know measure theory well (not me, unfortunately!), but in what sense would you like the $h_n$ to converge to $h$? –  Matt E Sep 28 '10 at 20:32
    
Pointwise. I'll add it. –  Jonas Teuwen Sep 28 '10 at 20:46
    
I've heard of a $\sigma$-finite measure, but what is a $\sigma$-finite sigma algebra? –  Byron Schmuland Sep 29 '10 at 0:42
    
I suspect that the answer is no, but that the required counterexample is pretty complicated. For what it's worth... –  Byron Schmuland Sep 29 '10 at 0:44
    
The measure is $\sigma$-finite, I'll correct it. –  Jonas Teuwen Sep 29 '10 at 10:09
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2 Answers

up vote 2 down vote accepted

In contrast to William, I think the answer is yes. I had to prove something similar once; here is a suitably adapted argument. It might be more complicated than necessary.

First, reduce to the case that $(\Omega_1, \mu_1), (\Omega_2, \mu_2)$ are finite measure spaces and $h$ is bounded. Let $\mathcal{P}$ be the set of all functions on $\Omega_1 \times \Omega_2$ of the form $F(x,y) = f(x) g(y)$ where $f$ is bounded and measurable and $g$ is simple; such functions can be written in the form you seek. Let $\mathcal{Q}$ be the linear span of $\mathcal{P}$; these functions can also be written in the desired form. Let $\mathcal{L}$ be the closure of $\mathcal{Q}$ in $L^1(\mu_1 \times \mu_2)$ and let $\mathcal{L}_b$ be the bounded functions from $\mathcal{L}$.

Now $\mathcal{L}_b$ is a vector space which is closed under bounded convergence (by the dominated convergence theorem), contains the constants and contains $\mathcal{P}$. $\mathcal{P}$ is closed under multiplication and contains all functions of the form $1_{A \times B}$ with $A \subset \Omega_1$, $B \subset \Omega_2$ measurable; the collection of such sets $A \times B$ generates the product $\sigma$-algebra. By the functional version of the Dynkin $\pi$-$\lambda$ theorem (references below), $\mathcal{L}_b$ contains all bounded measurable functions; in particular it contains $h$.

Since $h$ is in $\mathcal{L}$ which is the $L^1$ closure of $\mathcal{Q}$, there is a sequence $\mathcal{Q} \ni h_n \to h$ in $L^1$. Then some subsequence converges almost everywhere.

For the functional $\pi$-$\lambda$ theorem, see Theorem 8.2 of Bruce Driver's probability notes. A reference is also given to C. Dellacherie, Capacités et processus stochastiques, page 14.

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This looks good. I will wait if someone can come up with an easier argument to prove this before I accept it as answer. What was it that you had to prove? –  Jonas Teuwen Sep 29 '10 at 22:17
    
It was related to math.stackexchange.com/questions/1618/…. I wanted to prove that $L^p(X) \otimes L^p(Y) \subset L^p(X \times Y)$ was dense. –  Nate Eldredge Sep 29 '10 at 22:40
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The answer here is no. I am assuming you're trying to prove the following:

Given $\sigma$-finite measure spaces $\Omega_1, \Omega_2$, and $\Omega = \Omega_1\times\Omega_2$ carries the product measure, then if $h: \Omega\rightarrow \mathbb{R}$ is a measurable function, then $h$ is point-wise approximated almost everywhere by the sequence $h_n$ given by:

$$h_n(x,t) = \sum_{j=1}^{N_n}f_{j,n}(x)1_{F_{j,n}}(t)$$

Well, in general this is not the case since the sets $F_{j,n}$ will generally depend on $x$. If $\Omega_1$ is countable, the situation is not too bad: the family of sets $\{F_{j,n}(x): x\in \Omega_1\}$ is countable. In general, however, the situation is hopeless. I think using $\Omega_1 = \Omega_2 = \mathbb{R}_{\geq 0}$ and $h(x,t) = x^2 + e^t$, or something to that effect, should do as a counter example. Play with it.

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I'll try it if that works before I accept your answer. The $\sigma$-algebra is not necessarily countable. –  Jonas Teuwen Sep 29 '10 at 15:26
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$h(x,t) = x^2 + e^t$ is not a counterexample. We can easily approximate $e^t$ by functions of the desired form (take $f_{j,n}$ constant) and $x^2$ even more easily (take $f_{1,1}(x) = x^2$ and $F_{1,1} = \Omega_2$), and add them together. –  Nate Eldredge Sep 29 '10 at 18:42
    
@Nate: Indeed! I don't know what I was thinking. –  William Sep 30 '10 at 1:41
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