I want to check my proof. Let $\tau_a:E^{\Bbb R^n}\to E^{\Bbb R^n}$ such that $\tau_a f(x)=f(x-a)$ for all $x\in\Bbb R^n$ and any chosen $f\in E^{\Bbb R^n}$. Then the exercise says
Prove that if $\lim_{a\to 0}\|\tau_a f-f\|_\infty=0$ then there is some $g=f$ almost everywhere such that $g$ is bounded and uniformly continuous.
Here $\|{\cdot}\|_\infty$ is the seminorm of the space $\mathcal L_\infty(\Bbb R^n,\lambda_n,E)$, where $\lambda_n$ is the Lebesgue measure and $E$ is a Banach space. My proof:
The statement to be proved is not true in it generality. Let $f(x):=x$ (we choose here $E=\Bbb R^n$), then clearly $\|\tau_a f-f\|_\infty=|a|$ so $\lim_{a\to 0}\|\tau_af-f\|_\infty=0$, however $f$ is not essentially bounded, so I assume that we need the assumption that $f\in\mathcal L_\infty(\Bbb R^n,\lambda_n,E)$.
If $f\in\mathcal L_\infty(\Bbb R^n,\lambda_n,E)$ then clearly $f$ is essentially bounded, so there is some $g\in\mathcal L_\infty(\Bbb R^n,\lambda_n,E)$ such that $f=g$ a.e. such that $g$ is bounded.
Now suppose that $f$ is not uniformly continuous a.e., then for any chosen null set $N$ we have that $$ \exists \epsilon_0>0,\;\forall \delta>0,\;\exists x,y\in N^\complement:|x-y|<\delta\,\land\,|f(x)-f(y)|\ge\epsilon_0\tag1 $$ Then setting $|x-y|=|a|$ we find that $$ \exists \epsilon_0>0,\;\forall \delta>0,\;\exists a\in\Bbb R^n,\;\exists x_a\in N^\complement:|a|<\delta\;\land\;|\tau_a f(x_a)-f(x_a)|\ge\epsilon_0\tag2 $$ But then we find that $\|\tau_a f-f\|_\infty\ge\epsilon_0$ for arbitrarily small $|a|>0$ in any chosen subset of $\Bbb R^n$ with null complement, what contradicts the assumption that $\lim_{a\to 0}\|\tau_a f-f\|_\infty=0$, so we conclude that $f$ is also uniformly continuous a.e.
Now we need to show that if $f$ is uniformly continuous a.e. then there is some $g=f$ a.e. such that $g$ is uniformly continuous. Let $N$ the null set such that $f$ is uniformly continuous in $N^\complement$ and suppose that $\overline{N^\complement}\neq\Bbb R^n$, then $(\overline{N^\complement})^\complement$ is open and not empty, and because the Lebesgue measure is massive then $\mu((\overline{N^\complement})^\complement)>0$ what contradicts the completeness of the Lebesgue measure because $(\overline{N^\complement})^\complement\subset N$, hence $N^\complement$ is dense in $\Bbb R^n$.
Then by a previous result I know that $f|_{N^\complement}$ can be extended continuously on $N$, so we can conclude that such $g\in BUC(\Bbb R^n,E)$ exists.$\Box$