$\lbrace D^+ f < c \rbrace$ is measurable when $f$ is continuous

Question

Let $f: [a,b] \rightarrow \mathbb R$ be continuous and $$D^+f(x) = \limsup_{h \rightarrow 0^+} \dfrac{f(x+h) - f(x)}{h}.$$ Is the set $$\lbrace x \in [a,b] : D^+ f(x) < c \rbrace$$ Lebesgue measurable $\forall c \in \mathbb R$?

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I can rewrite $$ D^+f(x) = \limsup_{n} n(f(x+1/n) - f(x)). $$ But the $\limsup$ of measurable functions ($f$ is measurable) is still measurable, therefore the thesis.

Is this simple reasoning correct?

Answer 1

The supremum of a countable family of measurable functions is measurable. So is the infimum. Both follow from the fact that sub-level sets are countable unions (resp. intersections).

So, how to express $D^+f$ by means of such operations? Introduce $$ f_{h} = \frac{f(x+h)-f(x)}{h} $$ Take the supremum over all rational $h$ such that $0<h<1/n$. By virtue of continuity of $f$, this gives the same result as taking supremum over all real numbers in $(0,1/n)$.

Then take the infimum over $n$ to get $D^+f$.