Let $f\geq 0$ satisfy $\int_\mathbb{R} f < 1$. Let $f_n$ be the $n$ time convolution of $f$ by itself. Then I want to show $f_n \rightarrow 0$ a.e. as $n\rightarrow \infty$.
We can clearly obtain that $f_n \rightarrow 0$ in $L^1(\mathbb{R})$. But how to prove a.e. convergence?
There is a general result which says that an $L^p$ converging sequence has a sub sequence which converges almost everywhere. If you look at the proof (or see below), you will see that it actually holds for the whole sequence,if the convergence is "fast enough", for exponential.
In your example, let $\alpha =\| f \|_1 <1$. Since each $f_n$ satisfies $f_n\geq 0$, we can apply the monotone convergence theorem to get $$ \int \sum_n f_n \, dx =\sum_n \int f_n \, dx =\sum \|f_n\|\leq \sum \alpha^n <\infty. $$
But an integrable function is finite almost everywhere, so that $\sum f_n <\infty$ almost everywhere. In particular, $f_n \to 0$ almost everywhere.
Note that I used above that $\|f_n\|_{L^1}\leq \alpha^n$ in which follows from the standard estimate for convolution on $L^1$.