# absolutely continuous and increasing function

Im having a really hard time trying to solve this problem that appears in the book of Royden guy. Any help would be extremely appreciated.

Let $$f:[a,b]\rightarrow \mathbb{R}$$ an absolutely continuous and increasing function. Show that $$\lambda(f(A)) = \int_{A}f'd\lambda$$ for all $$A\subset [a,b]$$.

By now I've showed that $$\lambda(f(A))$$ sends null sets into null sets and $$G_{\delta}$$ into $$G_{\delta}$$. But Im pretty lost in the procedure of the integrals D:

Any help would be extremely appreciated! :)

• Start with the case $A=I$, an interval; then $f(A)$ is also an interval... Commented Jun 26, 2019 at 22:44

It suffices to prove this for Borel sets. Without loss of generality, $$a=0,\ b=1.$$ Set $$\mathscr S = \{A\subset\mathscr B([0,1]) : \lambda(f(A)) = \int_A f' \}.$$ Absolute continuity of $$f$$ implies that $$f(b)-f(a)=\int^b_af'$$ for all $$0 \le a\le b\le 1$$ and this in turn implies that $$\mathscr S$$ contains the intervals, and unions and intersections of finitely many intervals. Let $$\{A_n\}$$ be an increasing sequence of sets in $$\mathscr S$$ and set $$A=\bigcup_n A_n.$$ Then, $$\lim \int_{A_n} f'=\int_Af'$$ by the monotone convergence theorem. On the other hand, $$\lim \int_{A_n} f'=\lim \lambda(f(A_n))=\lambda f(A)$$ because $$\{f(A_n)\}$$ is increasing. Thus, $$A\in \mathscr S.$$ A similar argument shows that $$\mathscr A$$ is closed under decreasing sequences, and now by the monotone class theorem, $$\mathscr S=\mathscr B([0,1]).$$