Talk:Absolute continuity

I moved some portions of the old article in Signed measure. I have not had the time to add all Mathscinet and Zentralblatt references. Camillo 22:54, 29 July 2012 (CEST)

Could I suggest using $\lambda$ rather than $\mathcal L$ for Lebesgue measure since

• it is very commonly used, almost standard
• it would be consistent with the notation for a general measure, $\mu$
• calligraphic is being used already for $\sigma$-algebras

--Jjg 12:57, 30 July 2012 (CEST)

Why not? --Boris Tsirelson 13:21, 30 July 2012 (CEST)

Fine by me Camillo 14:08, 30 July 2012 (CEST)

Between metric setting and References I would like to type the following lines. But for some reason which is misterious to me, any time I try the page comes out a mess... Camillo 10:45, 10 August 2012 (CEST)

if for every $\varepsilon$ there is a $\delta > 0$ such that, for any $a_1<b_1<a_2<b_2<\ldots < a_n<b_n \in I$ with $\sum_i |a_i -b_i| <\delta$, we have \[ \sum_i d (f (b_i), f(a_i)) <\varepsilon\, . \] The absolute continuity guarantees the uniform continuity. As for real valued functions, there is a characterization through an appropriate notion of derivative.

Theorem 1 A continuous function $f$ is absolutely continuous if and only if there is a function $g\in L^1_{loc} (I, \mathbb R)$ such that \begin{equation}\label{e:metric} d (f(b), f(a))\leq \int_a^b g(t)\, dt \qquad \forall a<b\in I\, \end{equation} (cp. with ). This theorem motivates the following

Definition 2 If $f:I\to X$ is a absolutely continuous and $I$ is compact, the metric derivative of $f$ is the function $g\in L^1$ with the smalles $L^1$ norm such that \ref{e:metric} holds (cp. with )