For a measurable funtion on $[1,\infty)$ which is bounded on bounded sets define $a_n=\int_n^{n+1}f $. Is it true that $f$ is integrable over $[1,\infty)$ iff the series $\sum_na_n$ converges absolutely?
I think the answer is yes and I have already shown that if $f$ is integrable then $\sum_na_n$ converges absolutely. I am having trouble with the other direction.
Since $f$ is bounded on bounded sets and is measurable then $f$ is integrable on bounded sets and $\sum_{n=1}^{k}\int_n^{n+1}f=\int_1^{k+1}f$ by finite additivity. Also for each $k \in \Bbb N$ $\; \left| \int_1^{k+1}f \right| \le \sum_{n=1}^{k}\left|\int_n^{n+1}f \right| \le \sum_{n=1}^{\infty}\left|\int_n^{n+1}f \right| = M < \infty$ beacuse of the assumtion that $\sum_na_n$ converges absolutely. So since this is true for each $k$ then $\left| \int_1^{\infty}f \right| \le M$.
I cant figure out how to get $ \int_1^{\infty}\left|f \right| < \infty$ which is what we need to show $f$ is integrable. Any help is appreciated, thanks.
Might be reading this wrong, but consider a function that has $0$ integral on each interval $[n,n+1]$. For instance, consider $f(x)=-1$ on $[n,n+.5)$ and $f(x)=1$ on $[n+.5, n+1)$ for each $n \in \mathbb{Z}$. Then $\int_n^{n+1} f=0$ for each $n$, but $f$ is not integrable. The problem is $\vert a_n \vert = 0 \neq 1 = \int_n^{n+1} \vert f \vert$.