Limit Lebesgue measure of the preimage of a vanishing set (i.e. a set which tends to be empty) Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous real-valued function defined over a compact domain $[0, T] \subset \mathbb{R}$.
Let $c \in \mathbb{R}$ such that $\min_{x\in[0, T]} f(x) < c <\max_{x\in[0, T]} f(x)$.
It seems that the following result is true although I don't manage to prove it:
$$\lim_{h \to 0^+} \lvert f^{-1} \left(\ ]c, c+h[\ \right) \rvert = \lim_{h \to 0^+} \lambda\left( f^{-1} \left(\ ]c, c+h[\ \right) \right) = 0$$
where $\lambda(\cdot) = \lvert \cdot \rvert$ is the Lebesgue measure of a measurable set.
If someone comes up with a hint or a reference I would be very grateful!
 A: Hint: sets of the form $(c+\frac{1}{n+1},c+\frac{1}{n}]$ are disjoint. So are their preimages. Add up the measures of their preimages. Use facts about absolutely convergent series.
A: This is an alternative approach to @ChrisSanders's.
Define for some $\varepsilon>0$, $A_{\varepsilon, n}=(c,c+\varepsilon/n)$. We have $A_{\varepsilon,n}\supseteq A_{\varepsilon,n+1},\,\forall n$. Furthermore, $\cap_{n \in \mathbb{N}}A_{n,\varepsilon}=\emptyset$. Also, $f^{-1}(A_{\varepsilon,n})\supseteq f^{-1}(A_{\varepsilon,n+1})$ and $\cap_{n \in \mathbb{N}}f^{-1}(A_n)=f^{-1}(\cap_{n \in \mathbb{N}}A_n)=f^{-1}(\emptyset)=\emptyset$. If we can use continuity of measures, we have $\lambda(f^{-1}(A_{\varepsilon,n}))\to 0$. We can justify continuity of measures if there exists $\varepsilon>0$ s.t. $\lambda(f^{-1}(A_{\varepsilon,1}))<\infty$. In this case, this $\varepsilon$ exists because $f$ is continuous over $[0,T]$: just choose $\varepsilon=|c-\sup_{x \in [0,T]} f(x)|$. We then have $\lambda(f^{-1}(A_{\varepsilon,1}))\leq \lambda([0,T])=T$.
