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...



Shift operator that generates separable orbit
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 $0$.


