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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! :)

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

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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, $\mathscr S=\mathscr B([0,1]).$

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