# If $f\in L^1(\mathbb{R})$, then $\sum_{n\ge 1}f(x+n)$ Converges for a.e. $x$.

I am given that $f\in L^1(\mathbb{R})$ (i.e., $\int_{-\infty}^\infty\vert f \vert<\infty$). I would like to show that $$\sum_{n\ge0}f(x+n)\tag{*}$$ converges for almost every (a.e.) $x$, and I am looking for a technique that is general enough that can be applied to other sums (such as $\sum_{n\ge0}\frac{1}{\sqrt n}f(x-\sqrt n)$ - please feel free to include answers which contain other examples of sums which are convergent and defined in terms of $f$).

I was told that this problem can be solved by showing that a certain integral converges, but I don't see how to proceed that way.

What have I tried? It was a failed attempt based on a false assumption pointed out by a commenter, so it has been edited out. I could use some hints.

• It is not true that an integrable function is eventually decreasing - consider $f(x)=\frac{\sin x}x$ for a counterexample. Commented Feb 7, 2015 at 4:01
• @Jason: $\sin[x]/x$ is not in $L^{1}$. Commented Feb 7, 2015 at 4:16
• True. In that case use $f(x)=\frac{\sin x}{x^2}$. The same principle applies. Commented Feb 7, 2015 at 4:17
• Yes, but you should have quote a working example. That is all. Commented Feb 7, 2015 at 4:17
• Well, obviously I didn't mean to use one that didn't work... Commented Feb 7, 2015 at 4:19

You are correct in thinking that $\sum_{n=0}^\infty f(x+n)$ is absolutely convergent for almost every $x\in\mathbb{R}$. Let $k\in\mathbb{Z}$ be any integer. I claim $$\int_k^{k+1}\sum_{n=0}^\infty|f(x+n)|\ \mathrm{d}x<\infty.$$ Using the monotone convergence theorem we have \begin{align*} \int_k^{k+1}\sum_{n=0}^\infty|f(x+n)|\ \mathrm{d}x&=\sum_{n=0}^\infty\int_k^{k+1}|f(x+n)|\ \mathrm{d}x\\ &=\sum_{n=0}^\infty\int_{k+n}^{k+1+n}|f(y)|\ \mathrm{d}y\tag{$y=x+n$}\\\ &=\int_{k}^\infty|f(y)|\ \mathrm{d}y\\ &\le\int_{-\infty}^\infty|f(y)|\ \mathrm{d}y<\infty \end{align*} as claimed. Hence $\sum_{n=0}^\infty|f(x+n)|<\infty$ for almost every $x\in(k,k+1)$. Since $k\in\mathbb{Z}$ was arbitrary the result follows.