How to use equicontinuity? Rudin asked:
For $f\in L^{\infty}(\mathbb{R}^{1})$, define $f_{t}(x)=f(x-t)$. Assume that $$\lim_{t\rightarrow 0}|f_{t}-f|_{\infty}=0$$Prove that under these conditions there is a uniformly continuous function $g$ such that $g=f$ almost everywhere. 
The suggested strategy is to find a sequence of functions $h_{n}\rightarrow \delta$, and use $g_{n}=h_{n}*f$. It is easy to show that $g_{n}\rightarrow f$ almost everywhere(for example, a proof by contradiction or by dominated convergence theorem). However to bridge from $g_{n}$ to $g$ I encountered a difficulty; the classical Arzelà–Ascoli only works in compact spaces, and $g_{n}$ is not compact supported in general. Therefore it is not clear if $g_{n}\rightarrow g$ uniformly even if we only looking for a subsequence of the original sequence. 
 A: You do not need Arzelà–Ascoli here; the fact that $(g_n)$ converge uniformly can be shown directly. The key point is that $h_n$ should have small support when $n$ is large. This makes sure that the convolution integral involves only $f_t$ for small $t$.
$$ 
\sup |g_n-g_m| = \sup_x \left|\int h_n(t)f_t(x)\,dt- \int h_m(t)f_t(x)\,dt\right| \\ \le \epsilon + \sup_x \left|\int h_n(t)f(x)\,dt- \int h_m(t)f(x)\,dt\right| \\ = \epsilon + \sup_x \left|f(x)\cdot 1 - f(x)\cdot 1 \right| = \epsilon
$$
Here the second step is based on $\|f_t-f\|_\infty\le \epsilon$.
A: Here is another way: (I thought I had an elementary proof, but  I had to rely on a not entirely elementary result.)
Let $\phi(t) = \text{esssup}_x |f(x+t)-f(x)|$. We have $\lim_{t\to 0} \phi(t) = 0$.
Since $f \in L^\infty(\mathbb{R})$, we have $f \in L^1(K)$ for any bounded set $K$.
Let $f_n(x) = \frac{n}{2} \int_{-\frac{1}{n}}^\frac{1}{n} f(x+t) dt $ (and note that $f_n(x) = \frac{1}{2} \int_{-1}^1f(x+\frac{t}{n}) dt$). Since $f$ is essentially bounded, it follows that $f_n$ is continuous, and hence $f_n \in C([-M,M])$ for any $M>0$.
Suppose $n,m \geq N$, then we have
\begin{eqnarray}
|f_n(x)-f_m(x)| &\leq& \frac{1}{2} \int_{-1}^1 |f(x+\frac{t}{n})- f(x+\frac{t}{m})|dt \\
&\leq & \frac{1}{2} \int_{-1}^1 |f(x+\frac{t}{n})-f(x)|+|f(x)- f(x+\frac{t}{m})|dt \\
&\leq& \sup_{|t| \leq \frac{1}{N}}\phi(t)
\end{eqnarray}
Note that by assumption $\lim_{N \to \infty} \sup_{|t| \leq \frac{1}{N}}\phi(t) = 0$.
Choose $M>0$, then the above shows that $f_n$ is Cauchy in $C([-M,M])$, and hence converges to some $\hat{f} \in C([-M,M])$. It is clear that if $M'>M$, then is is clear that the limit coincides on $[-M,M]$, hence this procedure defines a continuous function $\hat{f}$ on all of $\mathbb{R}$.
Now for the not entirely elementary part: Since $f \in L^1(K)$ for any bounded $K$, it follows from the Lebesgue differentiation theorem that almost every $x$ is a Lebesgue point, ie, $\lim_{n \to \infty} f_n(x) = f(x)$ a.e. Since $\lim_{n\to \infty} f_n(x) = \hat{f}(x)$ everywhere, and $\hat{f}$ is continuous, we have the desired result.
