I'm proving that for $f: (\Omega, \mathcal{F}) \to [0, \infty]$ and a $\sigma-$finite measure $\mu$ on the $\sigma$-algebra $\mathcal{F}$, we have
$$\int_\Omega f d\mu = \int_{0}^\infty \mu \{f \geq t\} dt$$ where the latter integral is an improper Riemann integral.
In my proof, I got to:
$$\int_\Omega f d \mu = \int_{\mathbb{R}^+} \mu \{f \geq t\} \lambda(dt)$$ where the latter is a lebesgue integral, so it suffices for me to show that
$$\int_\mathbb{R^+} \mu \{f \geq t\} \lambda (dt) = \int_{0}^\infty \mu \{f \geq t\} dt$$
I know that if the latter Riemann integral exists, the function under the integrand is lebesgue-integrable and the integrals coincide.
But what if $\int_{0}^\infty \mu \{f \geq t\}dt = \infty$?
Best Answer
First of all, the formula holds only for non-negative functions $f \colon (\Omega,\mathcal{F}) \rightarrow [0,\infty]$. There is a theorem on the relation between Lebesgue- and Riemann-integrals. (For bounded function $g$ on a compact set $[a,b]$ it states: $g$ is Riemann integrable if and only if the set of discontinuity has a measure zero. And both integrals are identitcal.)
The Riemann integral exists on $[0,a]$, because $\mu\{f \ge t\}$ is a monotone function. Thus $$\tag{1}\int_0^a \mu \{f \ge t\} \, d \lambda(t) = \int_0^a \mu \{ f \ge t\} \, d t.$$ So if this term diverges, then $f$ is not Lebesgue-integrable and we have $\int f \, d \lambda = \infty$. On the other $\int f \, d \lambda < \infty$ implies that (1) is bounded. Therefore, as a monotone sequence, it is already convergent.