Let $f:[a,b] \to \mathbb{R}$ be absolutely continuous and increasing function.
Prove $$\int_A f'd\lambda=\lambda(f(A))$$ for all $A \subset [a,b]$ measurable.
My idea is,
$f$ absolutely continuous and increasing function. Then $f$ is integrable.
Define $$\mu:\mathbb{I} \rightarrow [0,\infty], \qquad \mu([a,b])=\int f'(x)dx.$$
Consider $$\mu([a,b])=f(b)-f(a)=\lambda([f(a),f(b)])=\lambda(f([a,b])).$$ Now, $\mu=\lambda \circ f\;$ in $\;\mathbb{I}\;$ and $\;\mu([a,b])<\infty.$
Then, by Caratheodory's Theorem, $\overline{\mu}$ is a unique extension.
$$\implies \overline{\mu}=\mu=\lambda \circ f \text{ in } \mathbb{B}.$$
Thanks so much for any help.
Going from intervals to measurable sets is not easy. So I am trying the following: Let $g(x)=x+f(x)$. Since $g$ is absolutely continuous and strictly increasing, hence one-to-one, it is easy to see that the result holds with $g$ in place of $f$. It remains only to show that $\lambda(g(A))=\lambda(f(A))+\lambda(A)$. I think I am struck here!