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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 $\mathbb{R}$.

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The last part looks false: if $f$ is non-negative and non-constant, then $g$ is non-integrable (and the Fubini argument goes down the drain). – D. Thomine Feb 21 '12 at 21:55
up vote 4 down vote accepted

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.

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