Shift operator that generates separable orbit

Question

Suppose, that $f$ is bounded measurable function, $T_h(f)(x) = f(x+h)$ is the shift operator. How to prove, that if the whole orbit $T_h(f):\, h\in\mathbb{R}$ has a dense, countable subset $T_{n_k}(f)$ (in $L^{+\infty}$ norm) then $f=g$ almost everywhere, where $g$ is continuous?

I ask for hints or ideas only :-)

I tried use something like modulus of continuity:

$\omega(f,\delta)=\mathrm{esssup}_{|x-y|<\delta} |f(x)-f(y)|$ and show that function $f$ must be uniformly continuous in the set of full measure, next extend from the dense set to entire $\mathbb{R}$. I didn't see, however, how to use separability...

Answer 1

The full proof:

In view of theorem D.A. Edwards (see "Translates of L∞ functions"), if translation operators are continuous at $0$, then $f$ is equal to a continuous function a.e.

Consider set $A= \{h: \|T_h f - f\| \leqslant \varepsilon \}$. For all $s\in\mathbb{R}$ we have from separability that $\|T_h f - T_d f\|\leqslant \varepsilon $ for appripriate $d \in D$ where $D$ is countable. Since $\|T_{h-d} f - f\| = \|T_h f - T_d f\|\leqslant \varepsilon$, we have $h-d \in A$ as well. Hence, $\mathbb{R} = \bigcup\limits_{d\in D} (d+A)$ and $\mu(A) > 0$. Note, that we used here Haar property of the Lebesgue measure. This property was also used by Edwards. By Steinhaus theorem, there is an open neighbourhood $U$ about $0$ in the set $A-A$. By triangle inequality we obtain $\|T_{h_1-h_2}f - Tf| = \|T_{h_1}f-T_{h_2}f\| \leqslant 2\varepsilon$. Hence $\|T_{h}f-f\|\leqslant 2\varepsilon$ whenever $h\in U$, what implies continuity of $T$ at $0$.