Rewriting an integral of an absolutely continuous function

Question

Let $(X,M, \mu)$ be a $\sigma$-finite measure space and $f: X \rightarrow [0, \infty)$ be measurable. For each $\alpha \ge 0$ Define $E_\alpha = \{ x \in X : f(x) > \alpha \}$ and $\lambda(\alpha) = \mu(E_\alpha)$.

Suppose that $\phi : [0, \infty) \rightarrow [0,\infty)$ is an increasing function which is absolutely continuous on [0,T] for every T $\in (0, \infty)$. Prove that for each $\beta \ge 0$,

$\int_{E_\beta}(\phi(f(x))-\phi(\beta))d\mu(x) = \int_\beta^\infty \phi '(\alpha)\lambda(\alpha) d\alpha$.

What I have so far:

$\int_{E_\beta}(\phi(f(x))-\phi(\beta))d\mu(x)=\int_{E_\beta} \phi(\beta)-\int_\beta^{f(x)} \phi ' (t) dt-\phi(\beta)d\mu(x)$ because $\phi$ is absolutely continuous.

$=-\int_{E_\beta}\int_\beta^{f(x)}\phi ' (t)dt d\mu(x) = \int_\beta^\infty\int_{E_t} \phi ' (t) d\mu dt$ (I'm not really sure why this is true?? Fubini-Tonelli allows us to switch the integrals, but I don't understand why the bounds change)

$=\int_\beta^\infty\phi ' (t) \mu(E_t)dt=\int_\beta^\infty \phi '(t)\lambda(t)dt$

Could someone please explain how the bounds change? Thanks!

Answer 1

We can write $$\begin{eqnarray} \int_{E_\beta}(\phi(f(x))-\phi(\beta))d\mu(x) &=&\int_{E_\beta}\left(\int_\beta^{f(x)}\phi'(t)dt\right)d\mu(x) \\ &=&\int_{E_\beta}\int_{\beta<t<f(x)}\phi'(t)dtd\mu(x) \\ &=&\int_{t>\beta}\int_{E_\beta\bigcap \{t<f(x)\}}\phi'(t)d\mu(x)dt\\ &=&\int_{t>\beta}\left(\int_{E_t} 1d\mu(x)\right)\phi'(t)dt\\ &=&\int_\beta^\infty \phi'(t)\lambda(t)dt, \end{eqnarray}$$ by Tonelli's theorem.

Answer 2

$f(x) >\beta$ and $\beta <t<f(x)$ iff $\beta <t<\infty$ and $f(x) >t$. Hence $x\in E_{\beta}$ and $\beta <t <f(x)$ iff $t >\beta$ and $x \in E_t$.