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Suppose $F:\mathbb{R}\to\mathbb{R}$ is a continuous function and $\delta>0$. Define for every $x\in\mathbb{R}$, $$ F_\delta(x):=\sup_{0

This is essentially a "simpler" version of this one. Playing around with the definitions of continuous functions and measurable functions, I still don't see how to go on.

By continuity of $F$, we have (since A continuous mapping is determined by its values on a dense set) $$ F_\delta(x)=\sup_{h\in(0,\delta)\cap\mathbb{Q}}\frac{F(x+h)-F(x)}{h}. $$

Thus one can reduce the supremum over a countable set for each $x\in\mathbb{R}$. But I don't see how this would help much.

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    http://math.stackexchange.com/questions/447762/a-sequence-of-measurable-functions-and-the-sup-lim-sup-of-them – 2017-01-01
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    @Michael Rozenberg: Thanks for your editing. – 2017-01-01
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    @Max: Thanks for the link. That is exactly the direction one is supposed to think. I somehow got stuck and Davide gives the complete answer. – 2017-01-01
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    @Jack You are welcome! – 2017-01-01

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You are almost done: since $\mathbb Q \cap (0,\delta)$ is countable, this set can be written as $\left\{q_n,n\in\mathbb N\right\}$. Consequently, $F_{\delta}(x)=\sup_{n\geqslant 1} g_n(x)$, where $g_n(x) :=\left(F\left(x+q_n\right)-F(x) \right)/q_n$. Now, the function $g_n$ is continuous hence measurable and a countable supremum of measurable functions is measurable.

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    That was the direction I was thinking about. I had some weird thought that the sequence $\{q_n\}$ might depend on $x$ which made me stuck. Thanks for your quick answer! – 2017-01-01
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Let $t_1 , t_2 , ...$ be a sequence of all rational numbers of the interval $(0, \delta ) $ and let $\Phi_n :\mathbb{R}\to\mathbb{R} ,$ $$\Phi_n (x) =\frac{F(x+t_n ) -F(x)}{t_n }.$$ The functions $\Phi_n $ are continuous and hence measurable and $$F_{\delta } (x) =\sup_n \Phi_n (x) ,$$ therefore $F_{\delta } $ is an measurable function.