What are the obstructions for a Henstock-Kurzweil integral in more than one dimension? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T22:11:22Z http://mathoverflow.net/feeds/question/34077 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34077/what-are-the-obstructions-for-a-henstock-kurzweil-integral-in-more-than-one-dimen What are the obstructions for a Henstock-Kurzweil integral in more than one dimension? Vladimir Sotirov 2010-08-01T01:12:40Z 2013-04-10T13:13:18Z <p>I have recently come across the book <a href="http://www.amazon.com/Kurzweil-Henstock-Integral-Differentials-Applied-Mathematics/dp/0824705351/" rel="nofollow">The Kurzweil-Henstock Integral and its Differentials</a> by Solomon Leader, in which, if I understand correctly, the HK integration process is modified in a way that makes it also work for dimensions higher than 1 (there's a proof of Green's theorem at the end). It has always been my impression that HK-integration doesn't extend to n dimensions, but truth be told, I don't actually know why. </p> <p>So my question(s) is (are):</p> <blockquote> <ol> <li>In what sense can the Henstock-Kurzweil integral not be extended to more than one dimension?</li> <li>Leader's construction via summants below seems very reminiscent of Jenny Harrison's work on chainlets. Are the two related?</li> <li>Does the relationship to measures from the one-dimensional case go both ways, i.e. every measure is a differential? Would this relationship be preserved in higher dimensions?</li> </ol> </blockquote> <p>Below I've summarized the key features of Leader's construction.</p> <hr> <p>A cell is a closed interval [a,b] in $[-\infty, \infty]$. A figure is a finite union of cells. A tagged cell in a figure K is a pair (I,t) where I is an interval contained in K, and t is an endpoint of I (according to Leader, the restriction of tags to be endpoints is key).</p> <p>A gauge is a function $\delta:[-\infty,\infty]\to (0,\infty)$. Every gauge associated to every point t a neighborhood '$N_\delta(t)$ which is $(t-\delta(t),t+\delta(t))$ for finite t, $[-\infty,-\frac1{\delta(-\infty)}]$ and $[\frac1{\delta(-\infty)},\infty]$ for the infinite points. This ensures that $N_\alpha(t) \subset N_\beta(t)$ if $\alpha(t)\leq\beta(t)$, and then we can define division of a figure K into tagged cells to be $\delta$-fine if for each tagged cell $(I,t)$ we have $I\subset N_\delta(t)$.</p> <p>Then where I understand Leader's theory to take a departure from the normal development, he defines a summant S to be a function on tagged cells, and then he constructs $\int_K S$ of a summant S over a figure K as the directed limit of $\sum_{(I,t)\in\mathcal{K}} S(I,t)$ over gauges $\delta$, where $\mathcal{K}$ are $\delta$-fine divisions of K (he actually defines a limit supremum and a limit infimum and works with those).</p> <p>Some summants are for example $\Delta([a,b])=b-a$ and $|\Delta|([a,b]=|b-a|$. Any summant S can be multiplied by a function by way of (fS)(I,t)=f(t)S(I,t), and any function can be canonically extended to a summant f\Delta.</p> <p>Where his theory really gets interesting is that he defines differentials as equivalence classes of summants under the equivalence relation S~T if $\int_K|S-T|=0$. From there he defines the differential of any function g by $dg=[g\Delta], where [S] is the equivalence class of the summant S.</p> <p>From this he calls a differential integrable if its representative summants S are integrable, and show that every integrable summant is of the form df where f is a function. Then absolutely integrable summants (ones such that |df| is integrable) give rise to measures. For example, the differential dx, where x is the identity function, corresponds to standard Lebesgue measure.</p> <p>A final point of interest is that the fundamental theorem of calculus can be formulated as$f\Delta=f'df$where f' is the usual derivative of f (which I actually think is a great way to motivate the definition of pointwise derivative in the first place).</p> <p>On to n-dimensional theory, though. An n-cell [a,b] consists of the parallelopiped with opposite vertices (a,a,a,...,a) and (b,b,b,b,...,b). A tagged n-cell (I,t) has tag one of the vertices, it is$\delta$-fine if it's diameter is less than$\delta(t)$. The$\Delta^(n)g$is given by the alternating sum$\sum_{t\in V_I} (-1)^{\mathcal{N}_I(v)}g(t)$, where V<sub>I</sub> are the vertices of I, and N<sub>I</sub>(v) are the number of coordinates of v that are the same as those of the tag t.</p> <p>Allegedly (Solomon doesn't give the details in his book), integration goes through, though some results regarding fundamental theorem and such allegedly do not. I am unclear on what happens to the relationship with measures.</p> <hr> http://mathoverflow.net/questions/34077/what-are-the-obstructions-for-a-henstock-kurzweil-integral-in-more-than-one-dimen/67519#67519 Answer by Anil P for What are the obstructions for a Henstock-Kurzweil integral in more than one dimension? Anil P 2011-06-11T16:11:41Z 2011-06-11T16:11:41Z <p>not many except need of a positive integrable function for fubini to hold</p> http://mathoverflow.net/questions/34077/what-are-the-obstructions-for-a-henstock-kurzweil-integral-in-more-than-one-dimen/67522#67522 Answer by Gerald Edgar for What are the obstructions for a Henstock-Kurzweil integral in more than one dimension? Gerald Edgar 2011-06-11T16:51:16Z 2011-06-11T16:51:16Z <p>There is a chapter on this (multi-dimensional gauge integral) in: W. Pfeffer, <em>The Riemann Approach to Integration</em> (Cambridge, 1993)</p> http://mathoverflow.net/questions/34077/what-are-the-obstructions-for-a-henstock-kurzweil-integral-in-more-than-one-dimen/127090#127090 Answer by Pedro Kaufmann for What are the obstructions for a Henstock-Kurzweil integral in more than one dimension? Pedro Kaufmann 2013-04-10T13:13:18Z 2013-04-10T13:13:18Z <p>This is reffering to question 1. </p> <p>There's a choice we have to make when defining a real-valued two-dimensional (or n dimensional) nonabsolute integral: would we rather have a class of integrable functions which include all divergences of differentiable functions, or would we want to have some sort of Fubini's Theorem working? This conflict was know from the early development of the HK integral and was also pointed out by Pfeffer in the book mentioned above. </p> <p>If we define the HK integral in the obvious way (call it the standard two dimensional HK integral), we get Fubini's theorem, but no fully general divergence theorem. Many authors made modifications on the definition of this integral, and an relatively satisfactory definition was given by Jarník, Kurzweil and Schwabik in "On Mawhin's approach to multiple nonabsolutely convergent integral", Casopis. Pest. Mat. There they defined the$M_1$-integral, which satisfies a fully general divergence theorem, and has a simple enough definition so we can prove some convergence theorems. It is shown though that this integral does not satisfy Fubini's theorem when the corresponding one-dimensional integrals are considered to be the HK integral. The original example from that paper can be used to show that, in a more general setting, an interval-based two-dimensional integral that satisfies a full divergence theorem will fail in some sense Fubini's theorem (see Proposição 2.5 <a href="https://dl.dropboxusercontent.com/u/54517768/myhomepage/artigos/phdfinal.pdf" rel="nofollow">here</a> if you are not afraid of reading in portuguese). </p> <p>Other problem that is frequently overlooked when defining interval-based two-dimensional nonabsolute integrals is that the integral can be sensitive to rotations, that is, we can get integrable functions such that a certain rotation of that function is not integrable. We have this unpleasant effect for the$M_1$-integral and even for the standard two-dimensional HK integral (see main theorem of "K teorii vícerozmerného integrálu", Casopis. Pest. Mat. by K. Kartak, if you are not afraid of reading czech, or Proposição 1.8 in the aforementioned Thesis, which is for the$M_1\$-integral but easily adaptable to the standard two-dimensional HK one). </p> <p>Then there is a new challenge: trying to define an integral which is not based on intervals but that still will be simple enough to prove convergence theorems. Kurzweil himself defined an integral where the domain is partitioned into sets with boundaries continuously differentiable by parts; it's a lot of trouble even to prove Saks' Lemma for this integral. See also <a href="https://dl.dropboxusercontent.com/u/54517768/myhomepage/artigos/KaufmannBianconi.pdf" rel="nofollow">this article</a> for an integral where we use triangular partitions. This integral satisfies many of the commonly desired theorems, but it is unknown to me for example if it satisfies a nice change of variables formula. </p>