# Talk:Absolute continuity

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## ...

Is there an example of a uniformly continuous function that is not absolutely continous? Albmont 17:43, 2 January 2007 (UTC)

Yes. The Cantor function, when restricted to the compact interval [0, 1], is a continuous function defined on a compact set, and is therefore uniformly continuous. However, is it not absolutely continuous, as the Cantor distribution is not absolutely continuous with respect to Lebesgue measure. Sullivan.t.j 18:10, 2 January 2007 (UTC)

I'd say we're missing the alternative characterisation of absolute continuity of measures here, the epsilon-delta one... Anyone can put it on? 189.177.62.204 01:21, 31 May 2007 (UTC)

That's not missing; it's in the article. Michael Hardy 01:25, 31 May 2007 (UTC)
I don't see it! There is an epsilon-delta definiton, but that one is for functions, not for measures. The one I mean is of the sort of: mu is abs. cont. w.r.t. nu if for every epsilon>0 there is a delta>0 such that if a set A satisfies mu(A)<delta then it satisfies nu(A)<epsilon. 189.177.58.19 22:21, 5 June 2007 (UTC)

Oh, OK. Go ahead and put it in. (But write either ν(A) < ε or $\nu(A) < \varepsilon\,$ or the like rather than nu(A)<epsilon.) Maybe I'll put it there if you don't, after I check a couple of sources....) Michael Hardy 22:29, 5 June 2007 (UTC)

you probably mean ν << μ iff for every ε > 0 there is a δ > 0 such that if a set A satisfies μ(A)< δ then it satisfies ν(A) < ε. the (<=) part is obvious. some finiteness assumption seems to be needed on ν, for a short proof of the converse:

suppose the ε-δ condition doesn't hold. so we have some ε and a sequence of sets An where ∑ μ(An) < ∞ and for every n, ν(An) > ε. Take the decreasing sequence Bm = Am ∪ Am+1 ... . Then μ(∩ Bm) = 0 but, if ν is finite, ν(∩ Bm) = lim ν(Bm) ≥ ε.
Your proof only works if you have finiteness assumptions on $\nu$, because only if at least one of the $B_{n}$ satisfies $\nu(B_{n}) < \infty$, your are allowed to compute $\nu(\bigcap B_{n})$ as $\lim_{n\rightarrow\infty}\nu(B_{n})$!
Let me give a counterexample. Let $\lambda$ be Lebesgue measure on the Borel-sets $\mathcal{B}$ of the real line. Then define (this is in a sense $\mu := \infty \cdot \lambda$): $\mu:\mathcal{B}\rightarrow\left[0,\infty\right],A\mapsto\begin{cases} 0, & \text{if }\lambda\left(A\right)=0\\ \infty, & \text{if }\lambda\left(A\right)>0.\end{cases}$. We shall prove, that $\mu$ is indeed a measure. Clearly $\mu\left(\emptyset\right)=0$. Now let $\left(A_{n}\right)_{n\in\mathbb{N}}\in\mathcal{B}^{\mathbb{N}}$ be a sequence of disjoint sets. We consider two cases:
• $\lambda\left(\bigcup_{n=1}^{\infty}A_{n}\right)=0$. Then we have $0\leq\lambda\left(A_{i}\right)\leq\lambda\left(\bigcup_{n=1}^{\infty}A_{n}\right)=0$ for alle $i\in\mathbb{N}$ and therefore by definition of $\mu$: $\sum_{n=1}^{\infty}\mu\left(A_{n}\right)=\sum_{n=1}^{\infty}0=0=\mu\left(\bigcup_{n=1}^{\infty}A_{n}\right).$
• $\lambda\left(\bigcup_{n=1}^{\infty}A_{n}\right)>0$. Then not all $A_{i},i\in\mathbb{N}$ can be of $\lambda$-measure zero, since otherwise we would have $\lambda\left(\bigcup_{n=1}^{\infty}A_{n}\right)=\sum_{n=1}^{\infty}\lambda\left(A_{n}\right)=\sum_{n=1}^{\infty}0=0,$ a contradiction. So let $i_{0}\in\mathbb{N}$ with $\lambda\left(A_{i_{0}}\right)>0$. Then we have again by definition of $\mu$: $\infty=\mu\left(\bigcup_{n=1}^{\infty}A_{n}\right)\geq\sum_{n=1}^{\infty}\mu\left(A_{n}\right)\geq\mu\left(A_{i_{0}}\right)=\infty,$ and therefore equalitiy.
Furthermore, by definition of absolute continuity and by definition of $\mu$ (we have $\mu\left(A\right)=0$ if $\lambda\left(A\right)=0$ for $A\in\mathcal{B}$), we see that $\mu$ is absolutely continuous with respect to $\lambda$. We will now prove that for $\varepsilon:=1>0$ there is no $\delta>0$ such that $\mu\left(A\right)<\varepsilon=1$ if $\lambda\left(A\right)<\delta$. To this end let $\delta>0$ be arbitrary and choose $A_{\delta}:=\left(-\frac{\delta}{4},\frac{\delta}{4}\right)\in\mathcal{B}$. Then $0<\lambda\left(A_{\delta}\right)=\frac{\delta}{2}<\delta$, but $\mu\left(A_{\delta}\right)=\infty\geq1=\varepsilon$.
What we have to assume at least to make the statement true (and your proof work), is that $\mu(A) < \infty$ whenever $\lambda(A) < \infty$ or something in that direction. --Phoemuex (talk) 09:19, 11 April 2010 (UTC)
You are right; I was not careful. Two corrections are made. Boris Tsirelson (talk) 16:30, 11 April 2010 (UTC)

is finiteness necessary? Mct mht 09:23, 25 July 2007 (UTC)

Should I add some information on so-called AC* functions (absolutely continuous in the narrow sense)? (in definition the value |f(xk)-f(yk)| is replaced by osc[x_k,y_k]f )? Or it should be in a different article? And what about ACG and ACG* functions? Probably this would require considering absolute continuity of functions on an arbitrary set E in R, instead of an interval. --a_dergachev (talk) 09:25, 14 February 2008 (UTC)

This page ought to have some mention of the Lebesgue version of the Fundamental Theorem of Calculus. To my mind that's a major reason for being interested in absolutely continuous functions. 72.25.102.62 (talk) 01:58, 4 September 2008 (UTC)

## Lipschitz vs absolute

I am not too familiar with absolute continuity but it seems a little odd that it should be stronger than Lipschitz, as the lead paragraph currently claims. Could someone add a clarification? Katzmik (talk) 12:09, 27 October 2008 (UTC)

P.S. In fact, I see that under "properties" it says the opposite. The lead should be corrected. Katzmik (talk) 12:11, 27 October 2008 (UTC)

## Reverting

Something strange happened to the definition in the last two edits. Reverted. Boris Tsirelson (talk) 08:27, 18 June 2009 (UTC)

## Interval I in the definition

Am I correct in assuming that this interval does not have to be finite?

TomyDuby (talk) 04:19, 22 November 2009 (UTC)

## Split

This should really be two articles: Absolutely continuous function and Absolutely continuous measure. Currently the article has two self-contained sections about two relatively distinct subjects, both of which would seem to be worthwhile subjects for articles of their own. There seems to be no reason for this other than inertia. The current referencing style makes in unclear which source corresponds to which subject, so the references would reviewed to see which article they should be placed in.--RDBury (talk) 20:11, 18 February 2010 (UTC)

No, I believe, these two notions are very much related (as written in Sect.2.1). The sources could be better; I'll add more sources soon. And they will be sources that do describe both subjects (and their relation). True, the relation becomes weak when we turn to functions with values in metric spaces. This case could be treated in a separate section, for convenience of readers that are interested only in real-valued functions. Boris Tsirelson (talk) 20:11, 3 March 2010 (UTC)
Two books are added. I'll also add some inline citations (but not for metric spaces). Boris Tsirelson (talk) 20:28, 3 March 2010 (UTC)

## Explanation re a small edit on 7th March

Today, I made what should have been a very small (one character) edit to the bit on R-N theorem. I apologise that it took me 4 attempts to get mu and nu the correct way around in the text of my accompanying edit summary. Hence the multiple undos and redos whose sole purpose was to correct the edit summary, not the page itself. — Preceding unsigned comment added by 138.40.68.40 (talk) 21:25, 7 March 2013 (UTC)