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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.

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$f_n \to f$ 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 $D \setminus E$.

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

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  • $\begingroup$ Nice one. Thanks. $\endgroup$ – user2959071 Jan 17 '18 at 1:34

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