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 $0\leq a<b\leq 1$. Thank you.


1 Answer 1


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.

  • $\begingroup$ 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 $[a,b]\subset[0,1]$? ($\{a\}\subset[a,a+\frac{1}{n}]$) $\endgroup$
    – Mike
    Sep 26, 2020 at 5:31
  • $\begingroup$ Yes, that is right. @Mike $\endgroup$ Sep 26, 2020 at 5:54

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