Prove series converge for almost every $x$ 
Let $f\in L^p(\mathbb{R})$, $1<p<\infty$, and let $\alpha>1-\frac{1}{p}$. Show that the series $$\sum_{n=1}^{\infty}\int_n^{n+n^{-\alpha}} |f(x+y)|dy$$ converges for a.e. $x\in \mathbb{R}$.

The question came from old qualifying exam.
Although not specified, in the context of the test measure on $\mathbb{R}$ is always Lebesgue measure.
I define $q=(1-\frac{1}{p})^{-1}$. The integral can be estimated by Hölder's inequality $$\begin{align}\int_n^{n+n^{-\alpha}} |f(x+y)|dy&=\int_{n+x}^{n+n^{-\alpha}+x} |f(y)|dy\\&=||f||_1\\&\leq||f||_p||1||_q\\&=\left(\int_{n+x}^{n+n^{-\alpha}+x} |f(y)|^pdy\right)^{\frac{1}{p}}(n^{-\alpha})^{\frac{1}{q}}\end{align}$$
Here $||\cdot||_p$ means the norm on $L^p \left([n+x,n+n^{-\alpha}+x]\right)$.
Suppose we can prove the integral $\int_{n+x}^{n+n^{-\alpha}+x} |f(y)|^pdy$ decays faster than $n^{-(1+\frac{1}{q})}$ then the $n^{th}$ term should decay faster than $\frac{1}{n}$ and therefore converge. However this is the farthest I can get, any idea how to proceed?
 A: Consider the non-negative function $$F(x):= \sum_{n\geq1}\int_{n}^{n+n^{-\alpha}}|f(x+y)| \, dy$$ and notice that if we manage to prove that $F\in L^{1}(\mathbb{R})$ the problem is settled. Now to lets rewrite $F$ by making the change of variable $y=t+n$. We will simply get $$F(x)=\sum_{n\geq1}\int_{0}^{1}\chi_{[0,n^{-\alpha}]}(t)|f(x+t+n)| \, dt$$ Now calculating the integral of $F$ we get 
$$\int_{\mathbb{R}}F(x) \, dx = \sum_{n\geq1}\int_{\mathbb{R}}\int_{0}^{1}\chi_{[0,n^{-\alpha}]}(t)|f(x+t+n)| \, dt \, dx = \sum_{n\geq1}\int_{0}^{1}\chi_{[0,n^{-\alpha}]}(t)\int_{\mathbb{R}}|f(x+t+n)| \, dx \, dt $$ where the first step is justified by Beppo-Levi's theorem and the last step by Tonelli's theorem. Now using Holder's inequality we get $$\int_{0}^{1}\chi_{[0,n^{-\alpha}]}(t)\int_{\mathbb{R}}|f(x+t+n)| \, dx \, dt \leq \left(\int_{0}^{1}\chi_{[0,n^{-\alpha})}(t) \, dt\right)^{1/q}\, \left(\int_{0}^{1}\left(\int_{\mathbb{R}}|f(x+n+t)|\, dx\right)^{p} \, dt\right)^{1/p}$$ Now using Jensen's inequality on the convex function $u\mapsto u^{p}$ (since p>1) we have that $$\int_{0}^{1}\left(\int_{\mathbb{R}}|f(x+n+t)|\, dx\right)^{p} \, dt \leq \int_{0}^{1}\int_{\mathbb{R}}|f(x+n+t)|^{p}\, dx \, dt= ||f||_{p}^{p}$$ Where the last step follows from that the Lebesgue measure invariant under translation. Now simplifying we get that $$\int_{0}^{1}\chi_{[0,n^{-\alpha}]}(t)\int_{\mathbb{R}}|f(x+t+n)| \, dx \, dt \leq \frac{1}{n^{\alpha/q}} \, ||f||_{p}$$ Hence we finally arrive at $$\int_{\mathbb{R}}F(x) \, dx \leq ||f||_{p} \, \sum_{n\geq 1} \frac{1}{n^{\alpha/q}} < \infty $$ since $\frac{\alpha}{q} >1$ by assumption, thus proving the claim.
A: Suppose $f\ge0$ to save typing. Let $q$ be the conjugate exponent.
Now
$$\begin{aligned}\int_0^1\int_n^{n+n^{-\alpha}}f(x+y)\,dydx
&=\int_n^{n+n^{-\alpha}}\int_0^{1}f(x+y)\,dxdy
\\&=\int_n^{n+n^{-\alpha}}\int_y^{y+1}f(x)\,dxdy
\\&\le \int_n^{n+n^{-\alpha}}\int_n^{n+2}f(x)\,dxdy
\\&= n^{-\alpha}\int_n^{n+2}f(x)\,dx\end{aligned}.$$
So if $F(x)$ is that sum then $$\int_0^1F(x)\,dx\le\int fK,$$
where $$K=\sum_{n=1}^\infty n^{-\alpha}\chi_{(n,n+2)}.$$
The fact that $\alpha q>1$ shows that $K\in L^q$. So $\int_0^1 F<\infty$, hence $F(x)<\infty$ for almost every $x\in[0,1]$. Similarly for $x\in[k,k+1]$.
