Measure theory & Lebesgue integral question

[xeno_gear]

Homework Statement

Here's an old qualifying exam problem I'm a little stumped on:
Let $$(X,\mu)$$ be a $$\sigma$$-finite measure space and suppose $$f$$ is a $$\mu$$-measurable function on $$X$$. For $$t > 0$$, let
$$\[ \phi(t) = \mu(\{x \in X : |f(x)| < t \}). \]$$

Prove that
$$\[ \int_0^{\infty} \frac{\phi(t)}{t^2}\, dm(t) = \int_X \frac{1}{|f(x)|}\, d\mu(x), \]$$

under the assumption that the LHS is finite. We may assume the $$(\mu \times m)$$-measurability of $$\{(x,t) : |f(x)| < t\}$$.

Homework Equations

$$dm(t)$$ represents integration with respect to Lebesgue measure.
EDIT: I know $$(X,\mu)$$ technically isn't a measure space since the $$\sigma$$-algebra isn't specified, but that's irrelevant, and the way it's presented here is exactly how the problem is stated. Thanks!

The Attempt at a Solution

I let $$E_t = \{x \in X : |f(x)| < t\}$$ and used the fact that, in general, $$\mu(E) = \int_X \chi_E(x)\, d\mu(x)$$. Then:

$$\begin{align*}<br /> \int_0^{\infty} \frac{\phi(t)}{t^2}\, dm(t)<br /> &= \int_0^{\infty} \frac{1}{t^2} \mu(E_t)\, dm(t)\\<br /> &= \int_0^{\infty} \frac{1}{t^2} \int_X \chi_{E_t}(x) \, d\mu(x)\, dm(t)<br /> \end{align*}$$
but I'm afraid that's as far as I got. I'm sure I need to use Tonelli somewhere here since the hypotheses are satisfied (sigma finiteness and nonnegative measurable functions) but I'm not sure how to use it. I also tried to put some sort of Hölder inequality on the LHS but got nowhere fast. Any help would be appreciated.. thanks!

 

[micromass]

Well, where you ended up, use Fubini. After this, you only need to check that

$$\frac{1}{|f(x)|}=\int_0^{+\infty}{\frac{1}{t^2}\chi_{E_t}(x)dm(t)}$$

 

[xeno_gear]

Got it, thanks!