Why is Lebesgue integration taught using positive and negative parts of functions? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:21:55Z http://mathoverflow.net/feeds/question/25161 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25161/why-is-lebesgue-integration-taught-using-positive-and-negative-parts-of-functions Why is Lebesgue integration taught using positive and negative parts of functions? KConrad 2010-05-18T18:29:17Z 2011-07-09T14:42:21Z <p>Background: When I first took measure theory/integration, I was bothered by the idea that the integral of a real-valued function w.r.t. a measure was defined first for nonnegative functions and only then for real-valued functions using the crutch of positive and negative parts (and only then for complex-valued functions using their real and imaginary parts). It seemed like a strange starting point to make the theory dependent on knowledge of the nonnegative function case when this certainly isn't necessary for Riemann integrals or infinite series: in those cases you just take the functions or sequences as they come to you and put no bias on positive or negative parts in making the definitions of integrating or summing. </p> <p>Later on I learned about integration w.r.t a measure of Banach-space valued functions in Lang's Real and Functional Analysis. You can't break up a Banach-space valued function into positive and negative parts, so the whole positive/negative part business has to be tossed aside as a foundational concept. At the <em>end</em> of this development in the book Lang isolates the special aspects of integration for nonnegative real-valued functions (which potentially could take the value $\infty$). Overall it seemed like a more natural method.</p> <p>Now I don't think a first course in integration theory has to start off with Banach-space valued functions, but there's no reason you couldn't take a cue from that future generalization by developing the real-valued case in the same way Banach-space valued functions are handled, thereby avoiding the positive/negative part business as part of the initial steps. </p> <p>Finally my question: Why do analysts prefer the positive/negative part foundations for integration when there is a viable alternative that doesn't put any bias on which function values are above 0 or below 0 (which seems to me like an artificial distinction to make)? </p> <p>Note: I know that the Lebesgue integral is an "absolute" integral, but I don't see that as a justification for making the very definition of the integral require treatment of nonnegative functions first. (Lang's book shows it is not necessary. I know analysts are not fond of his books, but I don't see a reason that the method he uses, which is just copying Bochner's development of the integral, should be so wildly unpopular.) </p> http://mathoverflow.net/questions/25161/why-is-lebesgue-integration-taught-using-positive-and-negative-parts-of-functions/25163#25163 Answer by coudy for Why is Lebesgue integration taught using positive and negative parts of functions? coudy 2010-05-18T18:49:36Z 2010-05-18T18:49:36Z <p>In standard textbooks, the Lebesgue integral is first defined for f with values in $[0,\infty]$. Note that the value $\infty$ is allowed, and it is important that it can be taken even on a set of positive measure.</p> <p>You can see why for example in the one-line proof of the Borel-Cantelli lemma: </p> <p>let $\mu$ be finite, and $A_i$ measurable sets so that $\Sigma \mu(A_i)&lt; \infty$. Then a.e. x belongs to only a finite number of $A_i$.</p> <p>Proof: $\int \Sigma\ 1_{A_i} d\mu = \Sigma\ \mu(A_i) &lt;\infty$. The function $x\mapsto \Sigma\ 1_{A_i}$ is integrable, hence finite a.e.</p> <p>If you start with a set of functions $f$ with values in $[-\infty,\infty]$, you will have to assume that either $-\infty$ or $\infty$ is taken only on a set of zero measure (so as to prevent the $\infty -\infty$ problem), and if you want it to work both for $f$ and $-f$, you will have to assume that none of $-\infty$, $\infty$ is taken on a set of positive measure, thus ruling out the kind of argument as in the Borel-Cantelli lemma. Which is of course unbearable for an analyst, and even more for a probabilist. But perfectly sound from a functional analysis perspective.</p> http://mathoverflow.net/questions/25161/why-is-lebesgue-integration-taught-using-positive-and-negative-parts-of-functions/25164#25164 Answer by Bill Johnson for Why is Lebesgue integration taught using positive and negative parts of functions? Bill Johnson 2010-05-18T18:49:59Z 2010-05-18T18:49:59Z <p>You want $\int f d\mu$ to make sense for every non negative measurable function and employ the monotone convergence theorem when the limit is not integrable. </p> <p>But sometimes I have developed the Lebesgue integral for bounded measurable functions supported on sets of finite measure and introduced $L_1$ as the completion of this. Some books do it similarly--define the integral for bounded measurable functions supported on sets of finite measure--but then define $\int f d\mu$ for non negative measurable functions and take differences when it makes sense. That mixed approach seems a bit weird to me.</p> http://mathoverflow.net/questions/25161/why-is-lebesgue-integration-taught-using-positive-and-negative-parts-of-functions/25165#25165 Answer by Mark Meckes for Why is Lebesgue integration taught using positive and negative parts of functions? Mark Meckes 2010-05-18T18:51:16Z 2010-05-18T18:51:16Z <p>I haven't looked at Lang's book before, but after a quick skim on Google Books, I see that his approach is to define the integral of simple functions, then use a completion with respect to the $L^1$ seminorm. One reason to avoid this approach is that it requires more functional analytic sophistication than we usually want to assume when first developing Lebesgue integration. This is true even for real-valued functions. Reducing first to the case of nonnegative functions allows the reduction to simple functions to be done by more elementary means.</p> <p>Another point is that a rigorous treatment of <em>improper</em> Riemann integration usually <em>does</em> involve splitting into positive and negative parts.</p> http://mathoverflow.net/questions/25161/why-is-lebesgue-integration-taught-using-positive-and-negative-parts-of-functions/25168#25168 Answer by Fabrizio Polo for Why is Lebesgue integration taught using positive and negative parts of functions? Fabrizio Polo 2010-05-18T19:28:24Z 2010-05-18T19:28:24Z <p>It comes down to this: which do you want to be simpler to deal with?</p> <p>1) Functions which have integrals of infinite magnitude. 2) Integration of vector valued functions.</p> <p>Traditionally, infinite integrals are seen as a more immediate obstacle (${\mathbb R}$ is an infinite measure space so this difficulty shows up quite quickly.) Such integrals probably seem more obviously relevant to students at first. So texts usually develop integration with (1) in mind from the beginning. </p> <p>To me, integration of vector valued functions is a lot more natural, and playing with the extended reals seems like little more than convenient notational trickery. But this is hindsight. I'm quite sure that when I first learned integration I would have been much more concerned with problems caused by infinity.</p> http://mathoverflow.net/questions/25161/why-is-lebesgue-integration-taught-using-positive-and-negative-parts-of-functions/25172#25172 Answer by Andrea Ferretti for Why is Lebesgue integration taught using positive and negative parts of functions? Andrea Ferretti 2010-05-18T20:26:53Z 2010-05-18T20:26:53Z <p>Well, it is not true that Riemann integral and series avoid the distinction altogether.</p> <p>If you want to define improper Riemann integrals, you can follow two ways. Say you want to define $\int_a^b f(x) dx$, where $a, b$ are not necessarily finite and $f$ need not be bounded.</p> <p>Either you split $f$ into the positive and negative part, or you define it as a limit of the truncated functions on a truncated domain, something like</p> <p><code>$$\lim_{t \to +\infty} \lim_{s \to a^+} \lim_{r \to b^-} \int_s^r \max \{ \min \{ f(x), t \}, -t \} dx$$</code> But then the result, for functions not in $L^1$, depends on the way you choose to go to the limit.</p> <p>Exactly the same happens for series: for those which are not absolutely convergent, the result depends on the order of summation.</p> <p>The trick to consider $f_{+}$ and $f_{-}$ allows one to consider improper integrals, which may as well be infinite, and to declare that the integral of $f$ does not make sense in those cases where the order of the limits is relevant.</p> <p>Of course you know all this, but I post it as an answer since it would be too long for a comment, so that you can comment to explain why this reason is not enough for you.</p> http://mathoverflow.net/questions/25161/why-is-lebesgue-integration-taught-using-positive-and-negative-parts-of-functions/25193#25193 Answer by Carl Offner for Why is Lebesgue integration taught using positive and negative parts of functions? Carl Offner 2010-05-19T01:10:56Z 2010-05-19T01:10:56Z <p>It's really the difference between two kinds of completions:</p> <ol> <li><p>An order-theoretic completion. For this, it's easiest to start with non-negative functions, and have infinite values dealt with pretty naturally.</p></li> <li><p>A metric completion. For this, it's more natural to start with finite-valued signed simple functions.</p></li> </ol> <p>It's not exactly that simple -- historically, signed simple functions (well, actually, I think they used step functions) were used in an order-theoretic treatment by Riesz and Nagy. But I think this is a good way to look at the two ways of approaching this integral.</p> <p>And needless to say, these two approaches generalize in two different contexts. They are both interesting and illuminate somewhat different aspects of the Lebesgue integral, even on the real line. For instance, the order-theoretic approach leads quickly to results such as the monotone convergence and bounded convergence theorems, while the metric approach leads naturally to the topology of convergence in measure and completeness of the $L_p$ spaces.</p> http://mathoverflow.net/questions/25161/why-is-lebesgue-integration-taught-using-positive-and-negative-parts-of-functions/25212#25212 Answer by Sergei Ivanov for Why is Lebesgue integration taught using positive and negative parts of functions? Sergei Ivanov 2010-05-19T08:21:04Z 2010-05-19T08:21:04Z <p>The integration of nonnegative functions deserves its own chapter, just like nonnegative measures. It has more features than the general case and there are cases when you need exactly these features and do not need negative numbers.</p> <p>And it is so elegant: every measurable function has an integral, and the integration is uniquely characterized by 3 properties: the integral of $1_A$ is the measure of $A$; the integration is additive and satisfies the monotone convergence theorem.</p> <p>For example, let $f:X\to Y$ be a map and $\mu$ a measure on $X$. Then one has a push-forward measure $\mu'$ on $Y$ defined by $\mu'(A)=\mu(f^{-1}(A))$. The integration against $\mu'$ is given by the formula $\int_Y hd\mu' =\int_X (h\circ f) d\mu$. Why? Because the r.h.s. satisfies the above 3 axioms. This is trivial to check, and going through any explicit definition of the integral would be painful.</p> <p>So essentially one proves the formula first for step functions, then for nonnegative functions via monotone limits, and then the general case follows. It is a standard type of an argument. And it looks natural and obvious to a student who learned the integration the traditional way.</p> http://mathoverflow.net/questions/25161/why-is-lebesgue-integration-taught-using-positive-and-negative-parts-of-functions/67439#67439 Answer by Piero D'Ancona for Why is Lebesgue integration taught using positive and negative parts of functions? Piero D'Ancona 2011-06-10T12:56:39Z 2011-06-10T12:56:39Z <p>The question has been exhaustively answered, I'd like to add a remark. The way I view the Lebesgue integral is: to every positive measurable function you can associate a meaningful integral (i.e. stable by all natural operations and limit procedures), which might be infinite. Now if you have a sign-changing measurable function, you can assign an integral to its positive and its negative part. If one of the two is finite, then you can associate a meaningful integral to that function too. If both are infinite, there are ways to define an integral in some cases, but much less meaningful and stable; some natural operations become impossible to define or unstable in general. I find this a quite natural approach: in most theorems you can replace the assumption 'integrable' with 'whose integral is defined'. Also, it is pretty intuitive that if both areas above and below the $x$-axis are infinite, there is little point in defining the signed area of the whole region.</p> <p>Consider for instance the Fubini theorem (for $L^1$ functions) and its counterpart for positive functions, which we call the Tonelli theorem here in Italy. You can actually merge the two results and just say: if the integral of a measurable $f(x,y)$ is defined, then both iterated integrals are defined and give the same result as the double integral.</p> http://mathoverflow.net/questions/25161/why-is-lebesgue-integration-taught-using-positive-and-negative-parts-of-functions/69877#69877 Answer by Jinxin Xue for Why is Lebesgue integration taught using positive and negative parts of functions? Jinxin Xue 2011-07-09T14:42:21Z 2011-07-09T14:42:21Z <p>This will naturally give the result, a function is Lebesgue integrable iff it is absolutely integrable. This is distinct from the Riemann integral. This absolute integrability is important. Look at the Riemann rearrangement theorem you will see the point. A series convergent but not absolutely convergent can achieve arbitrary limit by rearranging the series. The Lebesgue integral is a kind of rearrangement compared with Riemann integral.</p>