Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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}

share|improve this question
1  
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

1 Answer 1

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 $m(B)$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.