For a bounded, measurable function $f$ defined on $[1,\infty]$, does the fact that $f$ is integrable imply that $\sum_n a_n$ converges? Consider $f$ a bounded, measurable function defined on $[1,\infty]$ and define $a_n := \int_{[n,n+1)} f$. Does the fact that $f$ is integrable imply $\sum_{n=1}^\infty a_n$ converges?
There has been a post regarding this question but I would like to argue that the argument
\begin{equation}
\sum^\infty_{n=1} a_n = \int^\infty_1 f
\end{equation}
made in that post is not correct.
My attempt was the following:
Since $f$ is integrable, $|f|$ is also integrable. Therefore, \begin{equation}
\int_{[1,\infty)} |f| = \lim_{k\to \infty} \int^k_1 |f| < \infty.
\end{equation}
Moreover, we have the following:
\begin{align}
\sum^\infty_{n=1} a_n &= \sum^\infty_{n=1} \int^{n+1}_n f \\
&\leq \sum^\infty_{n=1} |\int^{n+1}_n f| \\
&\leq \sum^\infty_{n=1} \int^{n+1}_n |f| \\
&=\lim_{k\to\infty} \sum^k_{n=1} \int^{n+1}_n |f| \\
\end{align}
However, I cannot proceed further from here. Can anyone help?
 A: We have
\begin{align}
\sum_{n = 1}^{\infty}|a_n| &\leq \sum_{n = 1}^{\infty}\int_{[n, n + 1)}|f| \\
&= \sum_{n = 1}^{\infty}\int |f| 1_{[n, n + 1)} \\
&= \int \sum_{n = 1}^{\infty}|f| 1_{[n, n + 1)} \\
&= \int |f|1_{[1, \infty)} \\
&< \infty.
\end{align}
The second equality is by MCT and additivity of the integral on nonnegative functions. Alternatively, you can argue directly using DCT that $\sum_{n = 1}^{\infty}a_n = \int f1_{[1, \infty)}$.
A: Since $f$ is integrable, you know that
$$
\int\limits_1^{k+1}|f|\,dx=\sum_{n=1}^{k}\int\limits_n^{n+1}|f|\,dx
$$
converges to $\int_1^\infty |f|\,dx<\infty$ as $k\to\infty$ (by monotonic convergence, extending $f$ by zero beyond $k+1$). Thus your last limit is finite and your series $\sum a_n$ is convergent
A: I do think that a substantial part of the issue here is terminological, or similar. But not a mysterious mathematical issue (beyond a standard sort of example).
Namely, first, indeed, convergence, or even absolute convergence, of the series $\sum_n \int_n^{n+1} f(x)\;dx$ proves nothing about any sort of convergence or integrability or absolute integrability of $f$. The example of $f(x)=\sin 2\pi x$ illustrates this, and it's not a pathology, etc.
On another hand, equally robustly, if "$f$ integrable" means absolutely integrable, then by various means $\int_1^\infty f = \sum_n \int_n^{n+1} f$.
If real-valued $f$ is "integrable" in a technical Lebesgue-theory sense, namely, that one or the other of its positive and negative parts is absolutely integrable, while the other may fail to be so... we still reach the conclusion.
If there are other contexts of interest to the questioner, I'd be happy to entertain them. But, beyond the basic counter-example to a too-naive appraisal of the situation, I do not think there's a pathology here.
