TITLE: If $\int_{\mathbb{R}}\vert f\vert<\infty$, then $\sum_{n=1}^\infty\vert n\int_n^{n+1/n}f(x+y)dy\vert<\infty$
QUESTION [5 upvotes]: We are given $\int_{\mathbb{R}}\vert f\vert <\infty$. Want to show that for almost every $x$,  $\sum_{n=1}^\infty\vert n\int_n^{n+1/n}f(x+y)dy\vert<\infty$.
I have two ideas. The first is to let $f_n=\int_n^{n+1/n}\vert f\vert$. I can show by DCT that $f_n\to0$ so for a given $\varepsilon$ there is $N$ so that $n\ge N \rightarrow f_n < \varepsilon$. So $\sum_{n=1}^\infty\vert n\int_n^{n+1/n}f(x+y)dy\vert =\sum_{n=1}^N\vert nf(x+y)\vert+\sum_{n=N+1}^\infty n\varepsilon$ 
however, the last term tends to infinity so either I'm using the wrong $f_n$ or DCT doesn't apply.
Another idea is to apply Cauchy Criterion. Choose $n>m>N$ then $$\vert n\int_n ^{n+1}f-m\int_m ^{m+1}f\vert \\\le \vert n\int_n ^{n+1/n} f\vert + \vert m\int_m ^{m+1/m} f\vert$$ and if I want to show these two terms go to zero, I bump into the same problem above ($n\cdot\varepsilon$) unless I find a better upper bound.
How can I find upper bound for $\int_n ^{n+1/n}\vert f\vert$ that allows $n\int_n ^{n+1}\vert f\vert\to0?$

REPLY [3 votes]: Hint: Without loss of generality, assume $f(x)\ge 0$.
Define
$$
\psi(y)=\left\{\begin{array}{}
n&\text{if }n\le y\lt n+1/n&\text{for some }n\in\mathbb{Z}\\
0&&\text{otherwise}\\
\end{array}\right.
$$
Then
$$
\begin{align}
\sum_{n=1}^\infty n\int_n^{n+1/n}f(x+y)\,\mathrm{d}y
=\int_\mathbb{R}\psi(y)f(x+y)\,\mathrm{d}y
\end{align}
$$
Consider
$$
\begin{align}
\int_k^{k+1}\int_\mathbb{R}\psi(y)f(x+y)\,\mathrm{d}y\,\mathrm{d}x
&=\int_k^{k+1}\int_\mathbb{R}\psi(y-x)f(y)\,\mathrm{d}y\,\mathrm{d}x\\
&=\int_\mathbb{R}\left(\int_k^{k+1}\psi(y-x)\,\mathrm{d}x\right)f(y)\,\mathrm{d}y\\
&=\int_\mathbb{R}\underbrace{\left(\int_{y-k-1}^{y-k}\psi(x)\,\mathrm{d}x\right)}_{\le2}f(y)\,\mathrm{d}y
\end{align}
$$