Let $\lambda_n$ be the n-dimensional Lebesgue measure on $\mathbb{R}^n$, $E$ a Banach space and $p \in [1,\infty]$. For an arbitrary function $f: \mathbb{R}^n \rightarrow E$ and $a \in \mathbb{R}^n$, define $\tau_a f : \mathbb{R}^n \rightarrow E$ by setting $$ \tau_a f(x) := f(x-a) \quad \forall x \in \mathbb{R}^n. $$ Then $\tau_a \mathcal{L}^p(\mathbb{R}^n,\lambda_n,E) = \mathcal{L}^p(\mathbb{R}^n,\lambda_n,E)$ and for $p < \infty$, translation is continuous, i.e. $$ \lim_{a \rightarrow 0} || \tau_a f - f ||_p = 0 \quad \forall f \in \mathcal{L}^p. $$ For $p = \infty$, this is not true, but in exercise X.4.2 (iii) in "Analysis III" by Amann-Escher, I found the following statement:
If $\lim_{a \rightarrow 0} ||\tau_a f - f||_{\infty} = 0$ for a given $f$, there exists a bounded, uniformly continuous function $g: \mathbb{R}^n \rightarrow E$ s.th. $f(x) = g(x)$ a.e.
There are no conditions for $f$ given in the exercise, so I tried to prove it assuming $f \in \mathcal{L}^{\infty}$, but I didn't have any good ideas. My problem is basically that $|\tau_a f - f| \leq || \tau_a f - f ||_{\infty}$ holds only a.e., so to get uniform continuity, one would have to take out a set of measure zero for every $a \in \mathbb{R}^n$, but $\mathbb{R}^n$ is uncountable.
Is the statement correct (taking $f \in \mathcal{L}^{\infty}$), and if yes, how can one prove it? If no, under which conditions for $f$ does it hold?