From: bruck@mtha.usc.edu (Ronald Bruck) Newsgroups: sci.math Subject: Re: Proof of a double integral Date: 18 Jan 1995 13:29:18 -0800 In article <3fjs0e$8jj@newsstand.cit.cornell.edu> pdt2@cornell.edu (Paul Thurston) writes: >In article <3fjphn$6mt@mtha.usc.edu>, bruck@mtha.usc.edu says... >egral of g(x,y) in any arbitrary domain: (x1,x2), (y1,y2) >> >>I'm not sure how one verifies the conclusion for step functions and then >>uses monotone convergence. > >For the step functions argument, check out the proof of Theorem 1.39 on >page 30 of Rudin's "Real and Complex Analysis", 3rd edition, 1987. > >Also, note that this is a general fact of about measurable functions on >abstract measure spaces; there is no real need to appeal to finite >dimensinal euclidean space structure. > >Finally, this discussion can be thought of as proof of the uniqueness >assertion of the Lebesque-Radon-Nikodym theorem. I have the 2nd edition, in which Theorem 1.39 deals with ***non-negative*** measurable functions such that \int_E f\,d\mu = 0 \implies f = 0 \mu-a.e. That's quite a standard and trivial result. But the function in this problem is NOT given to be non-negative. Nor are you told that \int_E f\,d\mu = 0 for all measurable sets E, which indeed would be the uniqueness assertion of Radon-Nikodym (and immediately implies f = 0 a.e. from the previous paragraph). You have a function whose integral is zero on EVERY RECTANGLE, but you ONLY know it on rectangles, not on all measurable sets! You are going to have to use the theorem on uniqueness of extension of a countably-additive signed measure on a ring to the sigma-ring generated by that ring. I simply don't see any way around it. Concerning the original problem, I don't see how, given a measurable function f, NOT NECESSARILY NON-NEGATIVE, such that \int_{[a,b]\times[c,d] f(x,y)\,dx dy = 0 for all a < b, c < d, one can get a step function, or even a simple function, with the same properties. (I assume you mean simple functions, not step functions. Step functions are usually taken to be finite linear combinations of characteristic functions of INTERVALS, whereas simple functions are finite linear combinations of characteristic functions of measurable sets.) Like I said, I occasionally give this problem in graduate real analysis courses. I don't give credit when the student assumes the function is non-negative. --Ron Bruck ============================================================================== From: bruck@mtha.usc.edu (Ronald Bruck) Newsgroups: sci.math Subject: Re: Proof of a double integral Date: 18 Jan 1995 16:41:01 -0800 In article <3fk4hq$dat@newsstand.cit.cornell.edu: pdt2@cornell.edu (Paul Thurston) writes: :In article <3fk17e$8v7@mtha.usc.edu:, bruck@mtha.usc.edu says... :: ::In article <3fjs0e$8jj@newsstand.cit.cornell.edu: pdt2@cornell.edu (Paul :Thurston) writes: :::In article <3fjphn$6mt@mtha.usc.edu:, bruck@mtha.usc.edu says... :::egral of g(x,y) in any arbitrary domain: (x1,x2), (y1,y2) :::: ::I have the 2nd edition, in which Theorem 1.39 deals with :***non-negative*** ::measurable functions such that \int_E f\,d\mu = 0 \implies f = 0 :\mu-a.e. ::That's quite a standard and trivial result. :: ::But the function in this problem is NOT given to be non-negative. : : :This is true. But the integral of g vanishes on every subrectangle. : : You have a function whose ::integral is zero on EVERY RECTANGLE, but you ONLY know it on rectangles, ::not on all measurable sets! :: :Yes it does! :It is a straightforward exercise (using the Principle of Good Sets) to :verify that the integral of g vanishes on every measurable set, since :the subrectangles generate the Borel sets. ... :Look the problem was broken up into two parts. The non-negative case is :easy. For the general case, the fact that the integral vanishes over :every measurable set implies that it vanishes on the set where the :fucntion is strictly negative. Similarly for its strictly postive part as :well. :Now use the easy case to see that g has vanishing POSITIVE AND NEGATIVE :parts. : I don't think we're disagreeing on this. As I said in my post, you have to use the uniqueness of extension theorem; since the rectangles generate the Borel field on R^2, one has two countably-additive non-negative set functions-- in this case, the integrals of the positive and negative parts of f--which agree on a ring (the set of finite unions of half-open rectangles), and therefore agree on the sigma-algebra generated by that ring. Apparently this is what you call the Principle of Good Sets. But that's the whole point; Theorem 1.39 of Rudin is the easy part, and can only be applied AFTER you appeal to the extension theorem. (I STILL don't see how simple functions and the monotone convergence theorem come in. They're not needed to prove Theorem 1.39.) "Have to use" is a little strong, since as I indicated, you can also use the differentiation theorem. But that's quite a cannon to use on this little problem. Of course, when the function is continuous, the problem is VERY easy. I used to give the one-dimensional version of this as a problem in calculus courses. That was before they were dumbed down, of course :-( --Ron Bruck