absolute continuity on $R^{n}$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T02:49:30Z http://mathoverflow.net/feeds/question/87157 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87157/absolute-continuity-on-rn absolute continuity on $R^{n}$ nikita2 2012-01-31T18:28:42Z 2012-02-01T03:15:52Z <p>I know the definition of <em>absolute continuity</em> if there is a function $f:(a,b)\rightarrow R$. I wonder what is an analogy of this concept if we have a function $f:A\rightarrow R$, where $A\subset R^{n}$ is open set.</p> http://mathoverflow.net/questions/87157/absolute-continuity-on-rn/87158#87158 Answer by Tapio Rajala for absolute continuity on $R^{n}$ Tapio Rajala 2012-01-31T18:52:39Z 2012-01-31T19:52:46Z <p>There is a generalization of absolute continuity for mappings $f \colon A \to \mathbb{R}^d$. This is called <em>$n$-absolute continuity</em>. It was introduced by Jan Malý in <em>Absolutely continuous functions of several variables</em>. J. Math. Anal. Appl. <strong>231</strong> (1999) 492$-$508.</p> <blockquote> <p>A mapping $f\colon A \to \mathbb{R}^d$ is called <em>$n$-absolutely continuous</em> if for every $\epsilon > 0$ there exists $\delta > 0$ such that for any disjoint finite family $\{B_i\}$ of closed balls in $A$ we have $$\sum_i \mathcal{L}^n(B_i) &lt; \delta \quad\Longrightarrow\quad \sum_i \left(\text{osc}_{B_i}(f)\right)^n &lt; \epsilon.$$</p> </blockquote> http://mathoverflow.net/questions/87157/absolute-continuity-on-rn/87198#87198 Answer by Vaughn Climenhaga for absolute continuity on $R^{n}$ Vaughn Climenhaga 2012-02-01T03:15:52Z 2012-02-01T03:15:52Z <p>I guess it may depend on exactly which property of absolutely continuous functions you think is most important to keep, or to put it another way, exactly which definition you prefer in one dimension. For me the most commonly useful property of absolutely continuous functions is that they map sets of Lebesgue measure zero to sets of Lebesgue measure zero.</p> <p>Pulling in roughly equal parts from my memory of real analysis and what <a href="http://en.wikipedia.org/wiki/Luzin_N_property" rel="nofollow">Wikipedia</a> and <a href="http://www.encyclopediaofmath.org/index.php/Luzin-N-property" rel="nofollow">EoM</a> tell me, the story seems to be that a function $f\colon [a,b] \to \mathbb{R}$ is absolutely continuous if and only if all three of the following hold (Banach–Zaretskii theorem):</p> <ol> <li>$f$ is continuous;</li> <li>$f$ is of bounded variation;</li> <li>the Luzin N property holds: if $E$ has Lebesgue measure $0$, then so does $f(E)$.</li> </ol> <p>Each of these 3 properties generalises to higher dimensions. The first and third are immediate; the second requires a <a href="http://en.wikipedia.org/wiki/Bounded_variation#BV_functions_of_several_variables" rel="nofollow">slightly different definition of variation</a> than in one dimension, but is a completely standard thing.</p> <p>Thus one could say that a function $f\colon \mathbb{R}^m \to \mathbb{R}^n$ is "absolutely continuous" if those three properties hold, and to me this seems a very reasonable generalisation of the usual definition. (Of course one could extend this to maps between smooth manifolds, where you also have a notion of zero volume.)</p> <p>This seems to be different from the definition that Malý uses in the paper Tapio Rajala referred to in his answer. From a quick glance at that paper, there seem to be a number of different generalisations out there, and this seems to be another example of the phenomenon wherein various notions that are distinct in higher dimensions happen to all coincide in the lowest-dimensional case, so that you can generalise some aspects of the familiar setting, but not all. Which generalisation is useful depends on what your purpose is.</p>