Prove that there exists an uniformly continuous $g$ such that $f = g$ a.e Suppose $f \in L^{\infty}(\mathbb{R})$, $f_{h}(x) = f(x+h)$, and 
$$
\lim_{h \rightarrow 0}||f_{h} - f||_{\infty} = 0
$$
Prove that there exists a uniformly continuous function $g$ on $\mathbb{R}$ such that $f = g$ a.e

Firstly, I used the theorem such that if $f$ is Borel measurable and integrable, then there exists continuous function $g$ having compact support. However, I can't show $f$ is integrable, and Borel measurable. 
And I wonder about a notation $||f_{h} - f||_{\infty}$ means $||f(x+h)-f(x)||_{\infty}$. 
Can anybody help?
 A: Given $h>0$ let
$$
f^h(x)=\frac{1}{2\,h}\int_{x-h}^{x+h}f(y)\,dy=\frac{1}{2\,h}\int_{-h}^{h}f(x-y)\,dy.
$$
Lebesgue's differentiation theorem implies that $\lim_{h\to0}f_h(x)=f(x)$ almost everywhere. Moreover the family $f^h$ is uniformly bounded, since $|f^h(x)|\le\|f\|_\infty$ for all $h>0$.
Let's prove now that $f^h$ is equicontinuous. Given $x,x'\in\mathbb{R}$
$$
|f^h(x)-f^h(x')|\le\frac{1}{2\,h}\int_{-h}^{h}|f(x-y)-f(x'-y))|\,dy.
$$
Let $\epsilon>0$ be given. By hypothesis there exists $\delta>0$ such that
$$
|x-x'|<\delta\implies|f(x-y)-f(x'-y))|\le\epsilon\quad\forall y\in\mathbb{R}.
$$
Thus, if $|x-x'|<\delta$, then $|f^h(x)-f^h(x')|\le\epsilon$.
The Ascoli-Arzela theorem implies that there is a sequence $h_n>0$ such that $f^{h_n}$ converges uniformly.
A: I stumbled upon a similar question in Exercise 8.4 of Folland's "Real Analysis".
As I struggled to find my mistake of mixing up convergence with respect to the uniform norm and $\lVert \cdot \rVert_\infty$, I believe it is worth to point out a few remarks. I will make use of Julián Aguirre's notation:

*

*Let $h>0$. $f^h$ is a uniformly continuous function. Indeed: given $x,y \in \mathbb{R}$;

$$
\lvert f^h(x)-f^h(y)\rvert
\leq
\lVert f \rVert_\infty
\int
\lvert
\chi_{(x-h,x+h)}
-
\chi_{(y-h,y+h)}
\rvert
d m.
$$
One may use the DCT to prove that
$$
\lim_{y \to x} 
\int
\lvert
\chi_{(x-h,x+h)}
-
\chi_{(y-h,y+h)}
\rvert
d m
=
0.
$$
Due to translation-invariance of the Lebesgue measure, we obtain uniform continuity.


*If $\{ h_n\}_n$ is a sequence of positive numbers such that $\lim_n h_n = 0$, then $\lim_n \lVert f^{h_n} -f \rVert_\infty = 0$. To see that, first prove that $(f^{h_n})_n$ is a Cauchy sequence in $L^\infty$ and $f^{h_n} \xrightarrow[n]{} f$ a.e.

Up until now, I did not use the hypothesis $\lim_h \lVert f_h - f \rVert_\infty = 0$! This hypothesis is only necessary to prove that $\lim_h \sup \lvert f_h - f\rvert=0$.
To check out that difference, one could work out the details of approaching Heaviside's function $H\colon \mathbb{R} \to \{0,1\}$ with the means $H^h$: the sequence one obtains is a sequence of uniformly continuous functions which converges in $L^\infty$ yet does not convege with respect to the uniform norm. (hence the non-continuous limit)
