# Prove $\nu$ is absolutely continuous to Lebesgue measure if and only if $f$ is absolutely continuous.

Let $$\nu$$ be a finite Borel measure on $$[0,1]$$. Define $$f : [0,1] \to \mathbb R$$ by $$f(x) = \nu ([0,x))$$. Prove $$\nu$$ is absolutely continuous to Lebesgue measure (\mu) if and only if $$f$$ is absolutely continuous.

This is my attempt with the questions I have on it:

($$\Rightarrow$$) Assume $$\nu << \mu$$. Then whenever $$E \subset [0,1]$$ and

$$\nu(E) < \delta \rightarrow \nu (E) <\epsilon \leq \delta$$ (is this correct?)

Let $$\{[a_k, b_k)\}_{k \in \mathbb N}$$ be a collection of half-open intervals (Can the collection be Borel only, not open?) such that $$\mu ([a_k,b_k)) <\frac{\delta}{2^k}$$ for all $$k$$. It is clear that $$\sum_{k} |b_k -a_k|=\sum_{k} \mu ([a_k,b_k)) < \delta.$$

Then

$$\sum_{k} |f(b_k) - f(a_k)| = \sum_k \nu([a_k,b_k)) < \sum_k \frac{\epsilon}{2^k} =\epsilon.$$

The other side is going to be similar. Is my work correct?

• By no means is the other side similar. Commented Jan 9 at 5:00
• @geetha290krm can you solve it for me! Commented Jan 9 at 13:12

Let $$f$$ be absolutely continuous. You may note that $$f$$ is continuous and this implies $$\nu \{x\}=0$$ for every $$x$$. This allows us to use open intervals instead of half open intervals.
Let $$\mu (E)=0$$ and $$\epsilon>0$$. Let $$\delta>0$$ be chosen as in the definition of absolute continuity. There exists an open set $$V$$ such that $$E \subseteq V$$ and $$\mu (V)<\delta$$ . We can write $$V$$ as disjoint union of open intervals $$(a_i,b_i), i\ge 1$$. For any $$N$$ the total length of $$(a_i,b_i), 1\le i\le N$$ is less than $$\delta$$, so $$\sum\limits_{k=1}^{N}|f(b_i)-f(a_i)|<\epsilon$$. This gives $$\nu (\bigcup\limits_{i=1}^{N} ((a_i,b_i))<\epsilon$$ or $$\sum\limits_{i=1}^{N}\nu((a_i,b_i))<\epsilon$$. Let $$N \to \infty$$ to get $$\sum\limits_{i=1}^{\infty}\nu((a_i,b_i))<\epsilon$$. Finally, $$\nu (E) \leq \nu (V) \leq \sum\limits_{i=1}^{\infty}\nu((a_i,b_i))\le \epsilon$$ and $$\epsilon$$ is arbitrary, so $$\nu (E)=0$$. .