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Suppose we have a measure $\mu$ on $(a,b]$ such that $\mu(a,b]=F(b)-F(a)$ where $F$ is non-decreasing, continuous function from the right,

Definition: A function $F$ is said to be absolutely continuous if for every $\epsilon$ there exist a $\delta$ such that for each finite collection $[a_i,b_i], i=1,...,k$, of nonoverlapping intervals, $\sum (b_i-a_i)<\delta$ implies $\sum |F(b_i)-F (a_i)|<\epsilon$

I'd like to prove that:

  • $F$ is absolutely continuous if and only if for each Borel set $A$, $\mu(A+x)$ is continuous in $x$.

where $A+x=\{a+x:a\in A\}$

Thank you

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  • $\begingroup$ What have you tried? For the first question, you can start by writing down the definitions of absolute continuity. The answer will follow. For the second question, try assuming that $\mu(A)=0$ for all lebesgue measurable $A$. $\endgroup$ – Alex R. Jun 13 '16 at 20:36
  • $\begingroup$ Thanks @AlexR. There are lots of definitions for a measure to be absolutely continuous (with respect to Lebesgue measure), and unfortunately I don't know how to deal with continuity of $\mu(A+x)$ at $x$. $\endgroup$ – Stat95 Jun 13 '16 at 20:58
  • $\begingroup$ Presumably the notation $A+x$ implies translating $A$ by $x$? $\endgroup$ – Alex R. Jun 13 '16 at 21:01
  • $\begingroup$ Can you proof the assertion for all $A$ of the form $(a,b]$? If yes then you can approximate any $A$ as a union of these sets. $\endgroup$ – Alex R. Jun 13 '16 at 21:13
  • $\begingroup$ Actually, I'm not good at measure theory and I need an elementary proof, if it is possible. $\endgroup$ – Stat95 Jun 13 '16 at 21:32
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One Direction:

I didn't see for a while exactly how @AlexR 's hint "Can you [prove] the assertion for all $A$ of the form $(a,b]$? If yes then you can approximate any $A$ as a union of these sets." led to a solution. So a much more detailed version, one of those things that's a proof for some readers and a hint for others:

For anyone who wants just an actual hint of a hint: The aha for me was realizing that we should do said approximation with respect to Lebesgue measure $m$, not $\mu$.

First, it's not true without some additional finiteness assumption somewhere. Let $A$ be the union of the $(n,n+1)$ for even $n$, let $B$ be the union of the $[n,n+1]$ for odd $n$, and define $\mu(E)=m(E\cap B)$. The corresponding $F$ is $Lip_1$, hence absolutely continuous. But $\mu(0+A)=0$ while $\mu(x+A)=\infty$ for $0<x<1$.

Exactly what the extra hypothesis really is nobody knows. I'm going to show it's true if $m(A)<\infty$; then below we see it's true for all $A$ if $\mu(\Bbb R)<\infty$.

Lemma 0. If $F$ is continuous then $\mu(x+(a,b])$ depends continuously on $x$. Also $\mu(\{a\})=0$.

Lemma 1. If $F$ is absolutely continuous then for every $\epsilon>0$ there exists $\delta>0$ such that $m(A)<\delta$ implies $\mu(A)<\epsilon$.

(Hint: If $m(A)<\delta$ then $A\subset\bigcup(a_j,b_j)$ with $\sum(b_j-a_j)<\delta$.)

Now suppose $F$ is absolutely continuous and $A$ is a Borel set. Assume that $m(A)<\infty$. For each $n$ choose $\delta_n$ as in Lemma 1, with $\epsilon=1/n$. Let $A_n$ be a finite union of intervals with $$m(A\Delta A_n)<\delta_n.$$

Now $\mu(x+A_n)$ depends continuously on $x$, and $$|\mu(x+A)-\mu(x+A_n)|\le\mu((x+A)\Delta(x+A_n))<1/n,$$so $\mu(x+A)$ depends continuously on $x$, by uniform convergence.

Below: Now assume that $\mu(\Bbb R)<\infty$. Define $$\mu_n(E)=\mu(E\cap[-n,n]).$$Then $\mu_n(x+A)\to\mu(x+A)$ uniformly. So we can assume that $\mu$ is supported in some compact set.

Suppose $A$ is a Borel set and $x_0\in\Bbb R$. There exists a bounded set $B$ such that $$\mu(x+A)=\mu(x+B)\quad(|x-x_0|<1).$$So the previous case shows that $\mu(x+A)$ is continuous at $x_0$.


The Other Direction

Now suppose that $\mu(x+A)$ depends continuously on $x$ for every Borel set $A$. Suppose that $m(A)=0$. Fubini's theorem shows that $$\int\mu(x+A)\,dx=0.$$Since the integral depends continuously on $x$ this shows that $\mu(A)=0$.

So $m(A)=0$ implies $\mu(A)=0$. It's not hard to deduce from this that for every $\epsilon>0$ there exists $\delta>0$ such that $m(A)<\delta$ implies $\mu(A)<\epsilon$, and it's not hard to show that that implies $F$ is absolutely continuous.

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  • $\begingroup$ Thanks, but it needs extra lemmas (lemma 0) which is hard for me to understand. I've added a definition to my post, could you please prove it using this definition. $\endgroup$ – Stat95 Jun 14 '16 at 10:40
  • $\begingroup$ @EhsanTavassoli Not sure which lemma you're talking about, since you mention lemma 0, which has nothing to do with absolute continuity. In any case, I want to leave something for you to do. Lemma 0 is trivial, and Lemma 1 is very easy (note there's a hint for Lemma 1). $\endgroup$ – David C. Ullrich Jun 14 '16 at 13:18

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