An application of Fubini-Tonelli Let $f$ be a nonnegative measurable function which is finite $\mu$ almost everywhere.  Suppose that $\mu(E_t) < \infty$ for all $t>0$, where $E_t = \{x: f(x)>t\}$. Let $\lambda$ be another measure, where $\lambda((a,b]) = \mu(E_a) - \mu(E_b)$ for all $a<b<\infty$.  It is true that $\lambda$ is a Lebesgue-Stieljes measure though it is not necessary to prove (though the proof is straightforward - right continuity follows from continuity from above which is applicable since $\mu(E_t)$ is finite.
Prove that $\int f\,d\mu = \int_{(0,\infty)} \mu(E_t)\,dm(t) = \int_{(0,\infty)} t\,d\lambda(t)$.
Here $m$ is Lebesgue measure.
I was able to successfully prove the first equality in the case where $f$ is a simple function, and unsuccessfully do anything else. I wasn't able to conclude anything about $f$ in general because it's not clear how any convergence theorem would be applied.  So maybe what I need to do is somehow prove the second equality in general and then use it with a convergence theorem?
Any tips appreciated. I will also post a solution if I find one.
Thanks.
 A: $$\int f\mathrm d\mu=\int\int_0^\infty \mathbf 1_{[f\gt t]}\mathrm dt\mathrm d\mu=\int_0^\infty\int \mathbf 1_{[f\gt t]}\mathrm d\mu\mathrm dt=\int_0^\infty\mu(E_t)\mathrm dt
$$
Note the $\mathrm dt$ at the end (instead of the undefined $\mathrm dm(t)$ in the question).
A: For the first integral, you have to use Tonelli:
$$
\int_{(0,\infty)}\mu(E_t)\,dm(t)=\int_{(0,\infty)}\int_{E_t}1\,d\mu(s)\,dm(t)=\int_{\mathbb R}\int_0^{f(s)}1\,dm(t),d\mu(s)=\int_{\mathbb R}f(s)\,d\mu(s).
$$
The only difficulty here is the change of the limits. The region we are considering is initially described as 
$$
0\leq t<\infty,\ \ s\in E_t\, (\mbox{i.e. }f(s)>t);
$$
so we re-think of this as
$$
s\in\mathbb R,\ 0\leq t< f(s)
$$
For the second equality we use the same kind of trick:
$$
\int_{(0,\infty)}\mu(E_t)\,dm(t)=\int_{(0,\infty)}\lambda((t,\infty))\,d\lambda(s)\,dm(t)\\
=\int_{(0,\infty)}\int_{(t,\infty)}1\,d\lambda(s)\,dm(t)=\int_{(0,\infty)}\,\int_{(0,s)}\,dm(t)\,\lambda(s)=\int_{(0,\infty)}\,s\,d\lambda(s).
$$
Here again we are writing the region $0\leq t<\infty$, $t<s$ as $0\leq s<\infty$, $0\leq t<s$. We also used that $f$ is finite almost everywhere, as this implies that $\mu(E_b)\to0$ as $b\to\infty$; this implies that $\lambda((t,\infty))=\mu(E_t)$. 
