TITLE: Convergence of $\int\limits_{\mathbb{R}}\sum\limits_{n=1}^\infty f(nx+n!)dx$
QUESTION [8 upvotes]: Can you help me with the following question?
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function with the property that 
$\int_{-\infty} ^\infty |f(t)|\,dt < +\infty$.   
Show that for almost all $x\in \mathbb{R}$ (with respect to Lebesgue measure) the series 
$\sum_{n=1}^\infty f(nx+n!)$ converges and the formula $g(x):=\sum_{n=1}^\infty f(nx+n!)$ defines a Lebesgue integrable function on $\mathbb{R}$.

REPLY [3 votes]: This statement is not true in general.
Take  a non-negative function $f$ such that $0<\int\limits_{\mathbb{R}}f(x)d\mu(x)<+\infty$. 
Then we have non-negative functions $g_n(x)=f(nx+n!)$. By Beppo Levi theorem we have a measurable non-negative function $g(x)=\sum\limits_{n=1}^\infty g_n(x)$ and moreover
$$
\int\limits_{\mathbb{R}}g(x)d\mu(x)=
\int\limits_{\mathbb{R}}\sum\limits_{n=1}^\infty g_n(x) d\mu(x)=
\sum\limits_{n=1}^\infty\int\limits_{\mathbb{R}} g_n(x) d\mu(x)
$$
Note that
$$
\int\limits_{\mathbb{R}} g_n(x) d\mu(x)=
\int\limits_{\mathbb{R}} f(nx+n!) d\mu(x)=
\int\limits_{\mathbb{R}} f(nx) d\mu(x)=
\frac{1}{n}\int\limits_{\mathbb{R}} f(x) d\mu(x)
$$
so
$$
\int\limits_{\mathbb{R}}g(x)d\mu(x)=
\sum\limits_{n=1}^\infty\frac{1}{n}\int\limits_{\mathbb{R}} f(x) d\mu(x)=
\int\limits_{\mathbb{R}} f(x) d\mu(x)\left(\sum\limits_{n=1}^\infty\frac{1}{n}\right)=+\infty
$$
Thus we see that $g$ is not Lebesgue integrable.