Suppose $f \in L^\infty(\mathbb{R})$, $f\_h(x) = f(x + h)$, and$$\lim\_{h \to 0} \|f\_h - f\|\_\infty = 0.$$Does there exist a uniformly continuous function $g$ on $\mathbb{R}$ such that $f = g$ almost everywhere?



Existence of a uniformly continuous function $g$ on $\mathbb{R}$ where $f = g$ a.e.?
Yes.


Let $\{\phi\_n\}$ be a sequence of standard mollifiers, so $\phi\_n$ is continuous, $\phi\_n \ge 0$, and $\int \phi\_n = 1$. I claim the sequence $\{f \ast \phi\_n\}$ is uniformly equicontinuous. For let $\epsilon > 0$ and choose $\delta$ so small that $|h| < \delta$ implies $\|f - f\_h\|\_{L^\infty} < \epsilon$. Then for any such $h$ and any $x \in \mathbb{R}$ we have 
$$\begin{align\*}|(f \ast \phi\_n)(x) - (f \ast \phi\_n)(x+h)| &= |((f - f\_h) \ast \phi\_n)(x)| \\ &\le \int |f(y) - f\_h(y)| \phi\_n(x-y)\,dy \\ &\le \|f - f\_h\|\_{L^\infty} \cdot \int \phi\_n(x-y)\,dy \\ &< \epsilon \cdot 1.\end{align\*}$$
Moreover, $\|f \ast \phi\_n\|\_\infty \le \|f\|\_\infty$. Passing to a subsequence if necessary, we have $f \ast \phi\_n \to f$ almost everywhere. By Arzela-Ascoli, on any compact interval $[a,b]$ there is a subsequence of $f \ast \phi\_n$ converging uniformly to some continuous function $g$, which must equal $f$ almost everywhere on $[a,b]$. By taking larger and larger intervals, and noting that $g$ is well defined, we get $g$ defined on all of $\mathbb{R}$, still having $g$ continuous and $f=g$ almost everywhere.


Now since $g$ is continuous, we have $\sup\_x |g(x) - g\_h(x)| = \|f - f\_h\|\_{L^\infty} \to 0$ as $h \to 0$. So in fact $g$ is uniformly continuous.


