How improper Riemann and Legesgue integral associated? I'm looking for answer to Relation between the Lebesgue integral the improper Riemann integral
.
$$
(L) \int_{a}^{b}f(x)dx \: \: \: \: (R)\lim_{\alpha \to a+}\int_{\alpha}^{b}f(x)dx
$$
I found following theorem:
Let $f$ be a nonnegative continuous function. If $f$ is improperly Riemann integrable then it Lebesgue integrable on $\left(a, b\right]$ and we have 
$$
\int_{a}^{b}f(x)dx = \lim_{\alpha \to a+}\int_{\alpha}^{b}f(x)dx
$$
but there is no proof. Can somebody explain how to proof above theorem?

This question arose after reading about the Lebesgue integral of a function of arbitrary sign. Could this be somehow related to this?
Also I found similar question
improper Riemann integral and Lebesgue integral but its for $\left(0, 1\right]$

For $f$ a non-negative function on $(a,b]$, $f$ is (improperly-)Riemann integrable if
\begin{align}
\sup_{[x,y]} \int_x^y f(s)\mathrm{d}s < + \infty
\end{align}
where sur supremum is taken over all compact subinterval of $(a,b]$
If $f$ is as above, then the functions $ \mathbb{1}_{[a+\frac{1}{n},b]} \times f $ pointwise converges to $f$, and increaslingly. By monotony, you can calculate the integral of the limit, and it is finite.
