Example that a measurable function $f$ on $[1,\infty )$ can be integrable when $\sum _{n=1}^{\infty }\int_{n}^{n+1}f$ diverges.

Question

I am seeking help in my attempt to formulate a proof to disprove the following.

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 integrable over $[1,\infty )$ if and only if the series $\sum _{n=1}^{\infty }a_{n}$ converges ?

I strongly suspect this to be false and could prove that if the series converged absolutely then $f$ is integrable over $[1,\infty )$ but i am unsure how to extend this to answer the question based on conditional convergence of the series.

Any help would be much appreciated.

Answer 1

Try $f(x)=(-1)^n/n$ for every $n\leqslant x\lt n+1$ and $n\geqslant1$. This function $f$ is not Lebesgue integrable. One can modify the example, first, to get $f$ Riemann integrable on compact sets but not Riemann integrable on $[1,+\infty)$, and second (regarding your I could prove paragraph), to get $\sum\limits_n|a_n|$ convergent and $f$ not Lebesgue integrable.

Edit: A second example is $f(x)=1$ if $n\leqslant x\lt n+1/2$ for some $n\geqslant1$ and $f(x)=-1$ otherwise. Then $a_n=0$ for every $n$ hence $\sum\limits_n|a_n|$ converges, $f$ is locally Riemann integrable but not Riemann integrable on $[1,+\infty)$ since $x\mapsto\int\limits_0^xf$ oscillates between $0$ and $1/2$, and $f$ is not Lebesgue integrable either since $|f|=1$ identically.