# Show that Riemann integral and Lebesgue integral coincide.

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 $$\int_{0}^\infty \mu \{f \geq t\}dt = \infty$$?

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.