Problem: Assume that a real-valued function $g$ is continuous and differentiable on an open interval containing $[0,1]$, an assume that there exist constants $c_1$ and $c_2$ such that $0 < c_1 \le g'(x) \le c_2$ for all $x \in [0,1]$. Prove that $\mu(g(E)) = \int_E g'(x)dx$ for every Lebesgue measurable set $E \subset [0,1]$.
My attempt: I've been able to prove the case where $E$ is an open interval, and $g$ is an increasing function. This case is obvious: $\mu(g((a,b))) = g(b)-g(a) = \int_a^bg'(t)dt$. When $E$ is an open set, $E$ can be written as a disjoint union of open intervals, so that case follows as well. However, I can't figure out how to generalize this to Lebesgue measurable sets and to non-increasing functions.
If $g$ is non-increasing, and $E$ is an open interval, we still have that $g(x) = g(a) + \int_a^x g'(t)dt$. But I can't seem to come with with an expression for even $\mu(g((a,b)))$ when $g$ is not continuous. Help would be much appreciated.