A real analysis qualifying exam problem I was doing a real analysis problem set when I get stuck on this problem.
$1<p<\infty$ $f\in L^p(\mathbb{R})$ $\alpha>1-\frac1p$ show that
$$\sum_{n=1}^\infty \int_{n}^{n+n^{-\alpha}}|f(x+y)|dy<\infty$$ for a.e. $x\in\mathbb{R}$.
I have tried integrating the series w.r.t. $x$ on a bounded interval and try to show that it is finite using Holder inequality. But it doesn't give the things I want. Any hint or suggestion on it?
 A: I think I've come up with a solution.
By Young's inequality $ab\le a^p/p+b^q/q$ and Jensen's inequality $(\int_X f d\mu)^p\le \int_X f^pd\mu$ where $\mu(X)=1$ we have
$$
\begin{aligned}
\sum_{n=1}^\infty \int_{n}^{n+n^{-\alpha}}|f(x+y)|dy&\leq \sum_{n=1}^\infty \frac1p\left(\frac{\int_{n}^{n+n^{-\alpha}}|f(x+y)|dy}{n^{-\alpha}}\right)^p+\sum_{n=1}^\infty\frac 1qn^{-\alpha q}\\
&\le \sum_{n=1}^\infty \frac1pn^\alpha\int_{n}^{n+n^{-\alpha}}|f(x+y)|^pdy+\sum_{n=1}^\infty\frac 1qn^{-\alpha q}\\
\end{aligned}
$$
By assumption, the second term converges. For the first term, we integrate w.r.t. $x$ on $[a,a+1/2]$
$$\begin{aligned}
&\int_a^{a+1/2}\left(\sum_{n=1}^\infty \frac1p{\int_{n}^{n+n^{-\alpha}}|f(x+y)|^pdy}n^{\alpha}\right)dx\\
=&\sum_{n=1}^\infty \frac1p\int_{0}^{n^{-\alpha}}\int_{a+n}^{a+n+1/2}|f(x+y)|^pdxdy\cdot n^{\alpha}\\
\le &\sum_{n=1}^\infty \frac1p\int_{0}^{n^{-\alpha}}\int_{a+n}^{a+n+n^{-\alpha}+1/2}|f(x)|^pdxdy\cdot n^{\alpha}\\
=&\sum_{n=1}^\infty \frac1p\int_{a+n}^{a+n+n^{-\alpha}+1/2}|f(x)|^pdx n^{-\alpha} n^{\alpha}\\
\le& C\|f\|_p^p 
\end{aligned}
$$
The conclusion thus follows.
