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Let $f:[a,b] \rightarrow [0,\infty]$ where be Lebesgue measurable and $S=\{(x,y) \in \mathbb R^2:x \in \mathbb R, 0 \le y \le f(x)\}$. How can we show that $S$ is a Lebesgue measurable subset of $\mathbb R^2$ and $M_2(S)= \int_{\mathbb R} fdM_1$ where $M_k$ denotes the Lebesgue integral on $\mathbb R^k$ for $k=1,2$. Here $a,b$ is in the extended real line.

I have proved this result using the concept of product measures and Fubini's Theorem. But I want to prove this without getting into product measures.

Thanks for any help.

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I can't type the question as a whole.Don't know what is the problem! –  Ester Feb 22 '13 at 19:05
    
I have got the entire question typed now. –  Ester Feb 22 '13 at 19:16
    
You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. –  Zev Chonoles Feb 22 '13 at 19:19
    
$M_2 := M_1\otimes M_1$ by its definition, so how would you like to avoid dealing with product measures? –  Ilya Feb 22 '13 at 19:28
2  
Well, you know that $\int_\Bbb R f\mathrm dM_1 = \lim_n \int_\Bbb R f_n\mathrm dM_1$ where $f_n\uparrow f$ are simple functions. Thus you only have to show that $M_2(S_n) = \int_\Bbb R f_n\mathrm dM_1 $ (easy), and that $S_n\uparrow S$, then the result follows from the continuity of measures. –  Ilya Feb 22 '13 at 19:38

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