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 $\left(0, 1\right]$
 A: Consider the sequence $g_n = \chi_{[\frac {1}{n},1]}.f$ and apply the monotone convergence theorem.I,of course,assumed the domain to be $[0,1]$ but you get what I mean.
Edit :- You can observe why improper Riemann and Lebesgue integrals might not always be the same.The improper Riemann integral,if you note,is actually the limit of the integrals of $\chi_{[r,b]}.f$ as $r\to a$.This is actually a limit of integrals,which might not always equal the integral of the limit(Indeed,those functions converge pointwise to $f$ which might not necessarily be Lebesgue integrable).However,when assumptions like non negativity or integrability of $f$ are made,we can conclude that they are equal using the monotone and dominated convergence theorems respectively.
A: 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.
