Improper Lebesgue integral for non negative function If $f$ is non negative and improper Riemann integrable on $[0, \infty)$, prove that  $f$ is Lebesgue integrable and $\int_0^{\infty}f\mathrm{d}u = \lim_{n\rightarrow \infty} \int_0^nf\mathrm{d}u$. I need the link where I can get more information about this question and about improper Lebesgue integral in general.
 A: I don't think there's a definition of "improper Lebesgue integrable": we have to make a difference between Riemman integrable functions and improper Riemman integrable functions since the latter is defined as a limit of integrals and not exactly as an integral.
However, to solve the problem, you can use the Monotone Convergence Theorem. Given a subset $E$ of $\mathbb{R}$ let $\chi_E:\mathbb{R}\to \mathbb{R}$ be the indicator function of the set $E$, that is,
$$
\chi_E(x)=
\begin{cases}
1 & \mathrm{if}\, x\in E\\
0 & \mathrm{if} \, x\not\in E\\
\end{cases}
$$
Now, note that by definition $\int_0^n f \, dx= \int f\chi_{[0,n]}\, dx$. Define the sequence of functions $(f_n)$ given by $f_n=f\chi_{[0,n]}$ and note that $f_n\to f$ as $n\to \infty$ and that for every $n$, $f_n\leq f_{n+1}$. Since $f$ is nonnegative, all the $f_n$ are nonnegative. Also, since $f$ is improper Riemman integrable, we should have that $f$ is Riemman integrable on each of the intervals $[0, n]$ and since Riemman integrability implies Lebesgue integrability
, we have that all of the $f_n$ are Lebesgue integrable. Hence, by the monotone convergence theorem,
$$
\lim\limits_{n\to\infty}\int f_n\, dx= \int f\, dx
$$
Since $\int f_n\, dx= \int_0^n f\, dx$  and the improper Riemman integral of $f$ is defined to be $\lim\limits_{n\to\infty}\int_0^n f\, dx$, then
$$
\int f\, dx= \lim\limits_{n\to\infty}\int_0^n f\, dx<\infty
$$
And by definition, this means that $f$ is Lebesgue integrable.
