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Let $f(t,x):(0,\infty) \times \mathbb{R}^d \rightarrow \mathbb{R}$ be a bounded and $\mathcal{B}([0,\infty))\otimes \mathcal{B}(\mathbb{R}^d)$-measurable function. Is the function $$g(t,x) = \int_0^t f(s,x) ds $$ a $\mathcal{B}([0,\infty))\otimes \mathcal{B}(\mathbb{R}^d)$-measurable function? Why?

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For $\omega \in \Omega$, what is $g(\omega)$? I believe $g = \int f \mathrm{d}\mu \in \mathbb{R}$. –  k.stm Oct 3 '12 at 6:36
    
The general form of the question makes no sense. What is $\int^{-1}$? –  Michael Greinecker Oct 3 '12 at 6:37
    
I have suppressed the second question while I'm trying to rephrase it. –  Rojas Azules Oct 4 '12 at 9:54

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I will answer your first question. Your second question is not well-defined, as some of the commenters have noted. Returning to your first question, without any assumptions whatsoever on the measure spaces, I think the answer is no, but under some very weak assumptions - in particular that the two measure spaces in question are $\sigma$-finite, and that the integral of the function in question is finite - this is part of the content of the Fubini/Tonelli theorems for integration on product measures. So for your first question, since you're considering $\mathbb{R}^n$ and Borel measures, then the answer is yes so long as the integral is finite.

To see this, we first prove a result about measurable sets. If $(X,\Sigma_1,\mu), (Y,\Sigma_2,\nu)$ are the measure spaces, with $A \in \Sigma_1$ and $B \in \Sigma_2$ of finite measure, then take $E = A \times B$. If $E(x)$ and $E(y)$ be the "slices" of $E$ given by fixing $x$ and $y$, respectively, then $\nu(E(x)) = 1_A(x)\nu(B)$ and $\mu(E(y)) = \mu(A) 1_B(y)$ are measurable in $X$ and $Y$, respectively. By $\sigma$-finiteness, then we can take increasing sequences of products $A_i \times B_i$ to show that the same result holds for $A,B$ measurable. So this proves the desired result for when $f$ is a characteristic function. Then, to obtain the general result for measurable $f(x,y)$ on $X \times Y$, we use the fact that $f$ can be approximated pointwise by a sequence of simple functions.

I haven't actually proven (only sketched) my claims above. For more details, you can see any analysis textbook on integration.

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But since $f$ is measurable and bounded is integrable, am I right? Therefore their integral is finite. Also, the Lesbegue measure is σ-finite on $\mathbb{R}^d$ and your idea would follow, isn't it? –  Rojas Azules Oct 3 '12 at 8:27
    
Yes, sorry, I guess I was simply answering a broader question than you had actually asked. However, for your example, boundedness isn't enough. For example, take $f(x) =1$. It's not integrable, but it's measurable. I have edited my answer to adjust for this. –  Christopher A. Wong Oct 3 '12 at 8:48

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