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I would love some help with this problem:

Let $(X,\mathcal F,\mu)$ be a measurable space and let $f:X\to[0,\infty)$ be a positive Lebesgue integrable function. Prove that $$\int_X fd\mu = \int_0^\infty \mu(\{x\in X : f(x) > \lambda\})d\lambda.$$

I understand why this works geometrically in two dimensions but am not really sure how to go about showing this formally. I believe a good method would be to show the equality for step functions and then somehow generalize, but I'm not sure exactly how to do this either.

I'd appreciate any help.

Thanks!

Sliingshot
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3 Answers3

2

You can write $$ f(x) = \int_0^{f(x)} \, dt = \int_0^{\infty} 1_{ \{y: f(y)>t \} }(x) \, dt, \tag{1} $$ because $1_{ \{y: f(y)>t \} }(x)$ is $1$ for $t$ between $0$ and $f(x)$ and $0$ otherwise. Now, you use Tonelli's theorem, which says that

If $f:X \times Y \to [0,\infty)$, then $$ \int_{X \times Y} f(x,y) \, d\mu(x) \times d\nu(y) = \int_X \left( \int_Y f(x,y) \, d\nu(y) \right) \, d\mu(x) = \int_Y \left( \int_X f(x,y) \, d\mu(x) \right) \, d\nu(x), $$ in the sense that the integrals all either exist have the same value, or all diverge.

In this case, take $Y=[0,\infty)$, $\nu$ normal Lebesgue measure, and then you have $$ \int_X f(x) \, d\mu(x) = \int_X \left( \int_0^{\infty} 1_{ \{y: f(y)>t \} }(x) \, dt \right) \, d\mu(x) \\ = \int_0^{\infty} \left( \int_X 1_{ \{y: f(y)>t \} }(x) \, d\mu(x) \right) \, dt \\ = \int_0^{\infty} \mu\{ y:f(y)>t\} \, dt, $$ using the equality (1), then Tonelli, and lastly the definition of the integral of a characteristic function.

Chappers
  • 67,606
1

The following more general identity holds:

For $ f \in L^p(X), \quad 0 < p < \infty $ we have:

$$ ||f ||_p^p = p \int_0^{\infty} \lambda^{p-1} d_f ( \lambda ) \text{d} \lambda $$

where $ \displaystyle d_f ( \lambda ) = \mu \left( \{x \in X : | f(x) | > \lambda \} \right) $

Proof: Using Fubini's theorem http://en.wikipedia.org/wiki/Fubini%27s_theorem we have:

$ \displaystyle p \int_0^{\infty} \lambda^{p-1} d_f( \lambda) d \lambda = p \int_0^{\infty} \lambda^{p-1} \int_X \chi_{ \{x: |f(x)| > \lambda \} } d \mu (x) d \lambda= \int_X \int_0^{|f(x)|} p \lambda^{p-1} d \lambda d \mu(x)= $

$\displaystyle = \int_X |f(x)|^p d \mu (x)= ||f||_p^p $

passenger
  • 3,793
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You can prove that the integrand on the right, let me call it $F$, is actually a Riemann integrable function on compact intervals. There are various ways to prove this, but they all hinge on the fact that $F$ is monotone.

As such you can write the right side as $\lim_{n \to \infty} I_n$ where $I_n$ is a Riemann sum for integrating $F \chi_{[0,n]}$. This Riemann sum is the Lebesgue integral of a certain simple approximation of $f$. (Essentially it is the simple approximation where the range of $f$ on $[m/n,(m+1)/n]$ is counted as $m/n$ for $m=0,1,\dots,n^2-1$ and the range of $f$ on $[n,\infty)$ is counted as $n$.) Now you just need to prove that this simple approximation converges in $L^1$.

Ian
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