# If $g \in L^1$ and $f_n \to f$ a.e. where $|f_n| \leq 1$, then $g*f_n \to g*f$ uniformly on each compact set.

If $g \in L^1(\mathbb{R})$ and $(f_n)$ is a sequence of measurable functions converging to $f$ a.e where $|f_n| \leq 1$, then $g * f_n \to g * f$ uniformly on every compact set where $g*f = \int_{\mathbb{R}} g(x-y)f(y)dy$.

Using Egoroff's theorem, the sequence converges almost uniformly on every compact set; however, I am having difficulty extending this to the entire set.

edit:

$f_n \to f$ almost uniformly on a set $D$ if, for every $\epsilon > 0$, there is a set $E$ such that $\mu(E) < \epsilon$ and $f_n \to f$ uniformly on $D \setminus E$.

• What do you mean by "converges almost uniformly"? – David C. Ullrich Jan 16 '18 at 23:50
• please clarify your question it does not correlate with the title – Guy Fsone Jan 17 '18 at 0:37
• Sorry. I'm not sure how it does not correlate. If you are referring to the edit, that was in response to the question by David. – user2959071 Jan 17 '18 at 0:40
• Or you meant $g\ast f_{n}\rightarrow g\ast f$ in $L^{\infty}(K)$ for every compact set $K$? – user284331 Jan 17 '18 at 0:41
• I understood the question as being, on every compact set $K$, $g * f_n \to g * f$ uniformly. – user2959071 Jan 17 '18 at 0:47

It's clear from dominated convergence that $f_n*g\to f*g$ pointwise.
Recall that $g\in L^1$ implies that $$\lim_{h\to0}\int|g(t)-g(t+h)|\,dt=0.$$
It follows that the sequence $(f_n*g)$ is equicontinuous. And pointwise convergence plus equicontinuity implies uniform convergence (at least on compact sets).