# Show that $\lambda$ is absolutely continuous w.r.t. the Lebesgue measure $\mu$

Let $$F(x)$$ be nondecreasing and absolutely continuous function on $$[0,1]$$ with $$F(0)=0$$ and $$F(1)=1$$. Let $$\lambda$$ be the measure on the Borel $$\sigma$$-field $$\mathcal{B}$$ s.t. $$\lambda([a,b])=F(b)-F(a)$$. Show that $$\lambda$$ is absolutely continuous w.r.t. the Lebesgue measure $$\mu$$.

Definition of absolute continuity of $$F(x)$$ in terms of closed intervals: for any $$\varepsilon>0$$, $$\exists\delta>0$$ s.t. for any finite collection of disjoint $$\{[a_k,b_k]\}$$'s with $$\sum_{k}|b_k-a_k|<\delta$$, $$\sum_{k}|F(b_k)-F(a_k)|<\varepsilon$$.

Here are some thoughts I have so far: I need to take a Borel subset $$E\subset[0,1]$$ s.t. $$\mu(E)=0$$, and need to show that $$\lambda(E)=0$$. However, here the measure $$\lambda$$ is defined for closed intervals $$[a,b]\subset[0,1]$$.

If it is the open interval $$(a,b)\subset[0,1]$$, I can use the fact $$\lambda(E)=\inf\{\lambda(U): U\supset E\text{ and U is open}\}$$ to construct $$\{U_j\}\downarrow E$$ with $$\lambda(U_1)<\delta$$ s.t. $$\lambda(U_j)\to\lambda(E)$$. Since $$U_i$$ can be expressed as countably disjoint union of open intervals $$\{(a_j^k,b_j^k)\}$$, by absolute continuity of $$F(x)$$, as $$|b_j^k-a_j^k|<\delta$$, $$\sum_{k=1}^{N}\left|\lambda(a_j^k,b_j^k)\right|\leq\sum_{k=1}^{N}\left|F(b_j^k)-F(a_j^k)\right|<\varepsilon$$ Let $$N\to\infty$$, thus$$|\lambda(U_j)|<\varepsilon$$, which implies $$|\lambda(E)|<\varepsilon$$. Let $$\varepsilon\to 0$$, done.

But how can I deal with the closed intervals defined here? Since I know $$(a,b)=\bigcup_{n=1}^{\infty}[a+\frac{1}{n},b-\frac{1}{n}]$$, and $$\lambda(E)=\sup\{\lambda(K): K\subset E\text{ and K is compact}\}$$, will these help? And on $$\mathcal{B}([0,1])$$, $$\sigma((a,b))=\sigma([a,b])$$ for $$0\leq a. Thank you.

Absolutely continuous functions are continuous. So $$\lambda (\{a\})=\lim \lambda ([a,a+\frac 1 n])=\lim [F(a+\frac 1n )-F(a)]=0$$ for every real numbers $$a$$. It follows that all intervals with end points $$a$$ and $$b$$ have the same $$\lambda$$ measure.
• Thank you. So I can argue that $\lambda(\{a\})=\lambda(\{b\})=0$ for any $0\leq a<b\leq 1$, together with my proof for $(a,b)\subset[0,1]$ to prove for any closed interval $[a,b]\subset[0,1]$? ($\{a\}\subset[a,a+\frac{1}{n}]$) – Mike Sep 26 '20 at 5:31