If F is a positive $L^1$ function on R. Define $F_n = F(x+n)$. Show that there is a subsequence of $F_n$ converging to zero almost everywhere.
If I can show that $F_n$ converges to 0 in measure or if I could show $F_n$ converges in L1 norm then in both cases I think I am done. but I do not know how to get started.
The key is to slightly modify a standard proof that convergence in measure implies pointwise almost everywhere convergence of a subsequence. Let $$A_n(k)=\{x\in(-k,\infty)\colon |F_n(x)|>2^{-k}\}.$$ For each $k\in\mathbb N$, there exists $N_k\in\mathbb N$ such that $\mu(A_n(k))<2^{-k}$ for all $n\geq N_k$. We may assume that $N_1\leq N_2\leq \cdots$. Because $$\sum_{k=1}^\infty\mu(A_{N_k}(k))<\sum_{k=1}^\infty 2^{-k}<\infty,$$ it follows from the Borel-Cantelli lemma that $$\mu\left(\limsup_{k\to\infty}A_{N_k}(k)\right)=0.$$ If $x\in\mathbb R$ is such that $F_{N_k}(x)$ does not converge to $0$ as $k\to\infty$, then $x$ must belong to infinitely many of the sets $A_{N_k}(k)$. In other words, the set of points at which $F_{N_k}$ does not converge to $0$ is contained in the null set $\displaystyle \limsup_{k\to\infty} A_{N_k}(k)$. Thus, $F_{N_k}$ converges to $0$ pointwise almost everywhere.