Let $f:[a,+\infty)\rightarrow\mathbb{R}$ be integrable in every compact set of $[a,+\infty)$.
Prove that
$\int_a^{+\infty}f(t)dt<\infty\Longleftrightarrow\forall\epsilon>0$
$\exists a_0>a:\forall a_2>a_1>a_0$
$|\int_{a_1}^{a_2}f(t)dt|<\epsilon$
Hello everyone, I'm struggling with this problem. The strategy I'm thinking of is that since $F(x)= \int_a^{x}f(t)dt$ converges for $x\rightarrow\infty$ this means that for all sequence $x_n\rightarrow\infty$, $F(x_n)$ is a Cauchy sequence, because if it wasn't the integral wouldn't converge.
I feel that the idea is right but I'm not getting anywhere because i find it difficult to put it as a rigorous proof and I don't know how to go on from this, any help is appreciated!
$\int_a^{+\infty} f(t)\; dt$ is an improper integral. In the context of Riemann integration, the standard way of defining it is $$ \lim_{b \to +\infty} \int_a^b f(t)\; dt$$ The improper integral converges if and only if this limit exists. (I don't like writing "the improper integral converges" as $\int_a^{+\infty} f(t)\; dt < \infty$, because what if it diverges to $-\infty$, but presumably that's what is meant here)
Now use the $\varepsilon-N$ definition of limit as $b \to +\infty$.