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Assume that $f:E \to [0,\infty]$ where $E \subseteq \mathbb{R}^n$ is a measurable set, and $f$ is $\mathbb{L}$-measurable. And use $x \in \mathbb{R}^n$ and $y \in \mathbb{R}$.

First I'm wondering why the subsets A and B stated below are measurable. $$A=\{(x,y) \in \mathbb{R}^{n+1} | 0 \le y < f(x), x \in E\}$$ $$B=\{(x,y) \in \mathbb{R}^{n+1} | 0 \le y \le f(x), x \in E\}$$

And then how can I conclude following equation for measure value?

\begin{align} \lambda(A)=\lambda(B)=\int_E f(x)dx & = \int_0^\infty \lambda(\{x \in E | f(x) >y\})dy \\ &= \int_0^\infty \lambda(\{x \in E | f(x) \ge y\})dy \end{align}

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What is $\mathbb{L}, \lambda$ and the relation between these two? Lebesgue measure? – user20266 Jun 12 '12 at 16:23
Hint: $A$ is the union of sets $A_r = \{(x,y) \in {\mathbb R}^{n+1} \vert \ 0 \le y < r < f(x), x \in E\}$ for rational $r$. – Robert Israel Jun 12 '12 at 18:37
Yes. L is lebesgue measure. Rational r is sueful for inequality which has not equality. Thanks. – wowhapjs Jun 13 '12 at 3:48
up vote 0 down vote accepted

If $f$ is Lebesgue measurable, then the function $\tilde{f}(x,y)=y-f(x)$ is also Lebesgue measurable. Then $A=\{y \geq 0\} \cap \{ \tilde{f}(x,y)<0\}$ and $B=\{y \geq 0\} \cap \{\tilde{f}(x,y) \leq 0\}$ are both measurable.

For each $x \in E$, the slices $A_x=\{y \in \mathbb{R} \, | \, (x,y) \in A\}$ and $B_x=\{y \in \mathbb{R} \, | \, (x,y) \in B\}$ both have measure $f(x)$ (since they are, resp., the segments $[0,f(x))$ and $[0,f(x)]$). Thus,

$m(A)=\int \chi_A(x,y) \, dx \, dy=\int_E m(A_x)\,dx=\int_E f(x) \, dx$.

And similarly for $m(B)$.

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Then, how can I show that $\int_E f(x)dx & = \int_0^\infty \lambda(\{x \in E | f(x) >y\})dy $? – wowhapjs Jun 13 '12 at 3:52
$\int_E f(x)dx = \int_0^\infty \lambda(\{x \in E | f(x) >y\})dy$? – wowhapjs Jun 13 '12 at 3:55

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