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The following question is taken from Royden's Real Analysis $4$th edition, Chapter $4,$ question $29:$

Question: For a measurable function $f$ on $[1,\infty)$ which is bounded on bounded sets, define $a_n=\int_n^{n+1} f$ for each natural number $n.$ Is it true that $f$ is Lebesgue integrable over $[1,\infty)$ if and only if the series $\sum_{n=1}^\infty a_n$ converges?Is it true that $f$ is Lebesgue integrable over $[1,\infty)$ if and only if the series $\sum_{n=1}^\infty a_n$ converges absolutely?

The same question was asked in MSE. However, the OP did not ask the first part of the question. So I am asking here.

My attempt: First statement is partial correct, as its converse is false.

If $\sum_{n=1}^\infty a_n$ converges, then $f$ may not be integrable. Let $f(x) = \frac{(-1)^n}{n}$ if $n\leq x<n+1.$ Then $a_n = \frac{(-1)^n}{n}$ and therefore $\sum_{n=1}^\infty a_n$ converges (by Alternating Series Test) but $f$ is not integrable as $$\int_1^\infty |f| = \sum_{n=1}^\infty\int_n^{n+1} |f| = \sum_{n=1}^\infty \frac{1}{n}=\infty.$$ However, if $f$ is integrable, then $\sum_{n=1}^\infty$ must converge, as $$\sum_{n=1}^\infty a_n = \int_1^\infty f \leq \int_1^\infty |f| <\infty.$$

For second statement, I can only prove its sufficiency but have no idea on its necessity. If $f$ is integrable, then $$\sum_{n=1}^\infty |a_n| = \sum_{n=1}^\infty \left| \int_n^{n+1} f \right| \leq \sum_{n=1}^\infty \int_n^{n+1} |f| = \int_1^\infty |f|<\infty.$$

Can anyone check my proof and give some hints on necessity of second statement?

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    $\begingroup$ If $f$ is improperly Riemann integrable then $\sum a_n$ is convergent and if $f$ is Lebesgue integrable the $\sum a_n$ is absolutely convergent. The converse implications do not hold as shown by Gono. $\endgroup$ – Jack D'Aurizio Dec 30 '17 at 16:55
  • $\begingroup$ The catch is that you only know $|a_n| \le \int_n^{n+1} |f|$. $\endgroup$ – copper.hat Dec 30 '17 at 18:44
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If you take $f(x) = \sin(2\pi x)$ then $a_n \equiv 0$ and $\sum_{n=1}^\infty a_n$ converges absolutely but $f$ is neither improperly Riemann-integrable nor Lebesgue integrable.

The rest of your proof is fine.

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