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Suppose $f$ is integrable on $R^d$. For each $\alpha>0$, let $E_a = \{x:|f(x)|>\alpha\}$. Then whether there stands this: $$ \int_{R^d}^\ |f(x)|dx = \int_0^\infty m(E_\alpha)d{\alpha} $$

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What's $m$? Presumable some kind of measure, but you should explain that. Or, if it's supposed to be the lebesgue measure on $\mathbb{R}^d$, write $\lambda_d$ instead. –  fgp Apr 30 '13 at 15:30
    
@fgp $m$ may well be any $\sigma$-finite measure on $\mathbb{R}^d $ –  smiley06 Apr 30 '13 at 15:48
    
@smiley06 Huh? If $m$ is the zero measure, then the RHS most certainly is $0$, yet the LHS is unaffected... –  fgp Apr 30 '13 at 15:57
    
@simley06 Oh, I see, you assumed that LHS was meant to be $\int \ldots dm(x)$. Then I agree. –  fgp Apr 30 '13 at 15:59

3 Answers 3

Yes it does. You need to take product measure of $\mathbb{R}^d\times (0,\infty) $ which is say $ m\otimes \lambda $ where $\lambda $ is the 1 dimentional lebesgue measure. Then let $ E = \{(x,\alpha)\in \mathbb{R}^d\times (0,\infty)\ |\ |f(x)|> \alpha\} $, then you have corresponding sections as the $\alpha $ sction of that is as given $ E_\alpha = \{ x\in \mathbb{R}^d\ |\ (x,\alpha) \in E\} $ and the $x$ section $ E_x = \{\alpha \in (0,\infty)\ |\ (x,\alpha) \in E\} = (0,|f(x)|) $ Then use fubini to conlude (in your notation $dm = dx$ and $d\lambda = d\alpha$ and observe $\lambda(E_x) = |f(x)| $) $$ (m\otimes \lambda) (E) = \int_{\mathbb{R}^d}\lambda(E_x)dm = \int_{(0,\infty)} m(E_\alpha)d\lambda $$ Which gives you $$ \int_{\mathbb{R}^d} |f(x)|dx = \int^\infty_0 m(E_\alpha)d\alpha $$

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I'm going to show this for $d=1$ and $f > 0$, for notational convenience. $\lambda$ denotes the lebesgue measure on $\mathbb{R}$. You then have $$ \int_\mathbb{R} f \,dm = \int_\mathbb{R} \lambda([0,f(x)]) \,dm(x) = \int_\mathbb{R} \int_{[0,f(x)]} \,d\lambda \,dm(x) \overset{(Fubini)}= \int_A \,d(\lambda\times m) $$ where $$ A := \left\{(x,y) \in \mathbb{R}^2 \,:\, 0 \leq y \leq f(x)\right\} = \bigcup_{y > 0} \left(E_a \times \{y\}\right) \text{ for } E_y = \left\{x \in \mathbb{R} \,:\, f(x) > y\right\} \text{.} $$ Using that you can continue with $$ = \int_A \,d(\lambda\times m) \overset{(Fubini)}= \int_{\mathbb{R}^+} \int_{E_y} \,dm \,d\lambda(y) = \int_{\mathbb{R}^+} m(E_y) \,d\lambda(y) \text{.} $$ For the two Fubini applications to be valid you need $\int_A \,d(\lambda\times m) < \infty$. For a $\sigma$-finite measure $m$ the existence of that integral is guaranteed by the existance of $\int_\mathbb{R} \int_{[0,f(x)]} \,d\lambda \,dm(x)$ and thus by the existence of $\int_\mathbb{R} f \,dm$.

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This follows from integrating with respect to the measure $m$ on $\mathbb R^d$ the leftmost and rightmost parts of the identity $$ |f(x)|=\int_0^{|f(x)|}\mathrm d\alpha=\int_0^\infty\mathbf 1_{\alpha\lt|f(x)|}\,\mathrm d\alpha=\int_0^\infty\mathbf 1_{x\in E_\alpha}\,\mathrm d\alpha. $$ Thus, Fubini for nonnegative functions on $\mathbb R^d$ yields $$ \int_{\mathbb R^d}|f(x)|\,m(\mathrm dx)=\int_0^\infty m(E_\alpha)\,\mathrm d\alpha. $$

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