Suppose $f$ is a locally bounded, nonnegative, and measurable function on $[1,\infty)$ and define $\displaystyle \int_{n}^{n+1}f$, $\,\,\forall n \in \mathbb{N}$. Then, is it true that $f$ is integrable on $[1, \infty)$ if and only if the series $\displaystyle \sum_{n=1}^{\infty}a_{n}$ converges absolutely?
If so, how would you prove it? I'm thinking the Monotone Convergence Theorem, but I'm not entirely sure how to put it into practice.
In one direction, $f$ integrable implies $\sum_{n=1}^{\infty}a_{n}$ converges absolutely, I'm thinking the following:
Suppose $f$ is locally bounded, measurable, and nonnegative on $[1,\infty)$, and also that $f$ is integrable over $[1,\infty)$.
Then, $\int_{1}^{\infty}|f|<\infty$, and define $A:=[n, n+1]$. Then, $f \chi_{A}=f\vert_{A}$.
Since $\int_{1}^{\infty}|f|<\infty$, $\int_{1}^{\infty}|f|\vert_{A}<\infty$ implies that $\int_{A}|f|< \infty$, and since $\int_{A}|f| = \int_{n}^{n+1}|f| = |a_{n}|$, $sum_{n=1}^{\infty}|a_{n}| = \sum_{n=1}^{\infty}\int_{n}^{n+1}|f|$.
And that was as far as I got.
Please help. Full proofs preferred.