Integral over some measure is the same as integrating over the measure of the preimages of some intervals

Question

Sorry for the somewhat confusing question. I'm currently working on this problem:

Let ($\Omega, \mathcal{A}, \mu$) be a σ-finite measure space and $f: \Omega \rightarrow \overline{\mathbb{R}}$ with $\overline{\mathbb{R}} = \mathbb{R} \cup \{\pm \infty\}$ non-negative and measurable function. Then $$\int_\Omega f d\mu = \int_0^\infty\mu(\{f \geq t\})dt$$ where $\{f \geq t\} = \{x \in \Omega : f(x) \geq t\}$

I already figured it out for simple functions and are now trying to prove the general case. I've got so far: Let $(\varphi_k) \nearrow f$ with $(\varphi_k) = \sum_{i=1}^{n^{(k)}} a_i^{(k)} \chi_{A_i^{(k)}}$

Then: $$\int_\Omega f d\mu = \int \lim_{k \rightarrow \infty} \varphi_k d\mu = \lim_{k \rightarrow \infty} \int \varphi_k d\mu = \lim_{k \rightarrow \infty} \int_0^\infty\mu(\{\varphi_k \geq t\})dt$$

Now define $t_0 := \sum_{i=1}^{n^{(k)}} a_i^{(k)} = \max_{x \in \Omega} (\varphi_k(x))$

$\varphi_k$ definitely doesn't get bigger than $t_0$ and only "reaches" $t_0$ if $x \in \bigcup_{i = 1}^n A_i^{(k)}$. Therefore

$$\int_0^\infty\mu(\{\varphi_k \geq t\})dt = \int_0^{t_0} \mu(\bigcup_{i = 1}^n A_i^{(k)}) dt = t_0 \mu(\Omega)$$ since we can assume without loss of generality that $\bigcup_{i = 1}^n A_i^{(k)} = \Omega$.

But I don't really see where to from here. Anyone have a clue? Especially why the σ-finity is necessary.

Answer 1

Apply monotone convergence (again) at $\overset{!}{=}$, then we're left to prove $\overset{!!}{=}$:

\begin{align}\int_\Omega f d\mu = \int \lim_{k \rightarrow \infty} \varphi_k d\mu &= \lim_{k \rightarrow \infty} \int \varphi_k d\mu\\ &= \lim_{k \rightarrow \infty} \int_0^\infty\mu(\{\varphi_k \geq t\})dt\\ &\overset{!}{=}\int_0^\infty\lim_{k \to \infty}\mu(\{\varphi_k \geq t\})dt\\ &\overset{!!}{=}\int_0^\infty\mu(\{f \geq t\})dt\\ \end{align}

Proving $\overset{!!}=$ is really fun, and I'd spare any details for now if you'd like to try it yourself. It is based both on $(\varphi_k) \nearrow f$ and on the property you wondered how to use.

To justify monotone convergence at $\overset{!}{=}$, ensure that for each $t\in\mathbb R^+$, $g_t(k)=\mu(\{\varphi_{k}\geq t\})$

1. is non-decreasing - follows immediately from monotonicity of $\varphi_k$ and additivity of $\mu$.
2. converges (to what?). This one will follow from the $\overset{!!}=$ step (almost everywhere, which suffices for the theorem).

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Here's a hint for $\overset{!!}=$ if you don't even know how to start:

Compute the difference of the terms under the integrals $$\mu(\{f\geq t\})-\lim_{k\to\infty}\mu(\{\varphi_{k}\geq t\})$$ as much as possible to the form $$\lim_{k\to\infty}\mu(\textrm{some set})$$ and then look at the $f>t$ and $f=t$ cases separately.

And a bit more, should you need it:

For a fixed $t$, show that the above limit equals $$\lim_{k\to\infty}\mu(\{f\ge t>\varphi_k\}).$$ Next, split the cases $f>t$ and $f=t$, writing the set inside as the disjoint union $$\{f>t>\varphi_k\}\cup\{f=t>\varphi_k\}.$$ Now measure both sets separately. As $k\to\infty$, show that the first set eventually vanishes. Hence you're only concerned with the second set, where you can drop the $>\varphi_k$ part (why?) and the limit as a whole, so what's left to compute is $$\int_0^\infty \mu(\{x:f(t)=x\})dt.$$

Finally,

Is it possible for $$\int_0^\infty \mu(\{x:f(t)=x\})dt$$ to be strictly positive? If yes, that'd mean that for uncountably many $t$, the term inside was positive, i.e. $t$ would have a non-zero measure preimage $f^{-1}(t)$. If $T$ is the (uncountable) set of such $t$s, then $$\bigcup_{t\in T}f^{-1}(t)\subseteq\Omega$$ would be an uncountable disjoint union of positive-measure sets. But $\Omega$ is $\sigma$-finite, hence a contradiction (why?)!