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Problem: Let $E \subset R^n$ be a Lebesgue measurable set and $f : E \to [0,\infty)$ be a Lebesgue measurable function. Suppose $$A = \{(x,y) \in R^{n+1} : 0 \le y \le f(x), x\in E\}.$$ Let $\lambda_1$ denote Lebsgue measure in $R^1$, $\lambda_n$ denote the Lebesgue measure in $R^n$, and $\lambda_{n+1}$ denote the Lebesgue measure in $R^{n+1}$.

(a) Show that the set $A$ is Lebesgue measurable on $R^{n+1}$.

(b) Show that $$\lambda_{n+1}(A) = \int_E f(x) d\lambda_n (x) = \int_0^\infty \lambda_n (\{x \in E : f(x) \ge y\}) d\lambda_1(y).$$

My attempt at a solution: I know that this is just a basic application of the Fubini-Tonelli theorem, but I can't seem to wrap my head around it for some reason. For (a), I know that the set $A$ is the "area under the curve." But I'm not sure how to show that this is measurable from Fubini Tonelli. For part (b), the first equality seems obvious, and I don't know how much proof is necessary. For the second equality, I have $$\int_E f(x)d\lambda_n(x) = \int_0^\infty \int_{R^n} [f(x) \cdot \chi_E(x)] d\lambda_n(x) d\lambda_1(y),$$ by the Tonelli theorem, since $f$ is non-negative. But I don't quite see how to pull off that inner integration to get the integrand we want.

I would really appreciate some hints/intuitive explanations to point me in the right direction. Thanks!

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    $\begingroup$ For (a) think of $A$ as the inverse image of $[0,\infty)$ under the (measurable!) map $(x,y)\to f(x)-y$ from $R^n \times[0,\infty)$ to $R$. (This map is the composition of the measurable map $(x,y)\to (f(x),y)$ with the continuous map $(z,y)\to z-y$.) For (b), it might be helpful to write $f(x)$ as $\int_0^{f(x)} 1\,dy$> $\endgroup$ – John Dawkins Sep 15 '15 at 22:48
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For $(a)$, notice that the result is obvious if $f$ is a simple function. Indeed the region under the graph of a non-negative simple function looks like the disjoint union of sets of the form $A_i \times [0,a_i]$, which are of course measurable. Now let $f$ be any non-negative measurable function and let $\{s_n\}$ be an increasing sequence of non-negative simple functions that converges to $f$ for almost every $x \in E$. To conclude it is enough to notice that, up to a set of measure $0$, the area under the graph of $f$ is the union of the areas under the graphs of the simple functions $s_n$. You should be able to write down all the details.

For $(b)$, as you mentioned in the question, all you need to do is to apply Tonelli's theorem to $$\int_{\mathbb{R}^{n + 1}}\chi_A\, d\lambda^{n+1}.$$ The two expressions that you are after are the two iterated integrals. I think it helps noticing that $$\chi_A(x,y) = \chi_E(x)\chi_{[0,f(x)]}(y).$$

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  • $\begingroup$ Ah, I see. Part (a) makes sense now, thank you! For part (b), the edit you just made really clarifies things. Thanks! $\endgroup$ – poppy3345 Sep 16 '15 at 3:09
  • $\begingroup$ I am glad you got it :) $\endgroup$ – Giovanni Sep 16 '15 at 3:10
  • $\begingroup$ @Giovanni In (a), why did you take $(s_n)$ to be an increasing sequence of non-negative simple functions that converges to $f$ for almost every $x \in E$? Doesn't it converge to $f$ on all of $E$? $\endgroup$ – samantha Jul 14 '16 at 0:37
  • $\begingroup$ @sammy123: You are right, the a.e. is superfluous in this case! (although not incorrect ;) ) $\endgroup$ – Giovanni Jul 14 '16 at 3:23

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