"An improper Riemann integral is Lebesgue integrable if it is absolutely convergent. "
I've seen this statement quite often, but always without proof. I'd think if we have something like:
$$ \int_{0}^\infty |f(x)|dx<\infty$$
there has to be a point $c\in\mathbb{R}$, where the integral doesn't "grow" anymore i.e.
$$ \int_{0}^\infty |f(x)|dx \leq \int_{0}^c |f(x)|dx<\infty$$
otherwise the integral could not be convergent ( similar as to how an infinite sequence has to be a zero sequence in order for the infinite sum to be convergent ).
But I'm having trouble formalising this or in general finding a proper proof. What is a simple way to prove this theorem?
Best Answer
If $|f|$ is Riemann integrable on any bounded interval this coincides with the Lebesgue integral. We have by the monotone convergence theorem,
$$\int_0^\infty |f(x)| \, dx = \lim_{c \to \infty} \int_0^c |f(x)| \, dx = \lim_{c \to \infty} \int_{[0,c]} |f| = \lim_{c \to \infty} \int_{[0, \infty)} |f| \chi_{[0,c]} = \int_{[0,\infty)}|f|$$
Here I use $\int_a^b g(x) \, dx$ to denote a Riemann integral and $\int_{[a,b]} g$ to denote a Lebesgue integral.