Convergence in $L^1$ 
Let $f \in L^1(R)$. Show that $\sum^{\infty}_{n=1} f(x + n)$ converges
  a.e.

Solution:
So, ultimately we are going to want $\sum^{\infty}_{n=1} f(x + n) \leq$ something in $L^1$ that converges to 0. This requires that I need to relate $\sum^{\infty}_{n=1} f(x + n)$ with something that satisfies the form of $L^1$, $\int f(x+n) dx$. What should I do to get started in that way?
 A: Let us show, e.g., that the series converges absolutely a.e. on $[0, 1]$. But this is obvious by noting that integral of $\sum_{n \in \mathbb{Z}} \int_{[0, 1]}|f(x + n)|dx$ is just the $L^1$ norm of $f$. 
A: The series $g(x):=\sum_{n\geq 1}|f(x+n)|$, which takes values in $[0,+\infty]$ a priori, is measurable as a (nondecreasing) pointwise limit of (nonnegative) measurable functions, the partial sums. By monotone convergence and change of variable
$$
\int_{(0,1)}g(x)dx=\lim_{n\rightarrow+\infty}\int_{(0,1)}\sum_{k=1}^n|f(x+k)|dx=\lim_{n\rightarrow+\infty}\sum_{k=1}^n\int_{(0,1)}|f(x+k)|dx
$$
$$
=\lim_{n\rightarrow+\infty}\int_{(1,n+1)}|f(x)|dx=\int_{(1,+\infty)}|f(x)|dx<\infty.
$$
So $g$ is integrable over $(0,1)$, which implies in particular that it is finite a.e. In other terms, the series $\sum_{n\geq 1}f(x+n)$ converges absolutely a.e. on $(0,1)$. 
Repeat the trick on every interval $(m,m+1)$, for $m\in\mathbb{Z}$. This proves absolute convergence a.e. on $\mathbb{R}$.
A: For each $a\in\Bbb Z$, give $[a,a+1)\times \Bbb Z_{>0}$ the product measure of Lebesgue measure and counting measure and use Fubini's theorem on the function $(z,n)\mapsto f(z+n)$.  This shows that the sum converges a.e. for $x\in [a,a+1)$.  Since $a$ was arbitrary the result follows.
