Prove that meas$(A)\leq 1$ Let $f:\mathbb R\to [0,1]$ be a nondecreasing continuous function and let $$A:=\{x\in\mathbb R : \exists\quad y>x\:\text{ such that }\:f(y)-f(x)>y-x\}.$$
I've already proved that: 
a) if $(a,b)$ is a bounded open interval contained in $A$, and $a,b\not\in A$, then $f(b)-f(a)=b-a.$
b)$A$ contains no half line.
What remains is to prove that the Lebesgue measure of $A$ is less or equal than $1$. I've tried to get estimates on the integral of $\chi_A$ but i went nowhere further than just writing down the integral. I'm not sure whether points a) and b) are useful, but i've reported them down for the sake of completeness so you can use them if you want.
Thanks everybody.

hint
(c) show that $A$ is open
(d) Any open subset of $\mathbb{R}$ is the union of countably many disjoint open intervals $\amalg (a_i,b_i)$. The Lebesgue measure is equal to $\sum (b_i - a_i)$. Now apply item (a) and the fact that $f$ is non-decreasing. 
