From: Ronald Bruck Subject: Re: Sequence of continous functions Date: Tue, 20 Feb 2001 13:08:23 -0800 Newsgroups: sci.math Summary: Pointwise limit of continuous functions on R has a point of continuity In article <3A929B9F.599FB4AC@city.dk>, Michael Knudsen wrote: :Hello ! : :Consider af sequence {f_n} of continous functions f_n:R->R, and assume :that f_n(x) -> f(x) for n -> oo for all x in R. It can be shown that f :in continuos at at least one point x_0 in R. Does anybody have any ideas :about how to show this? Whoof. Perhaps there is a simple proof for the EXISTENCE of a point where f is continuous, but if so, I don't know it. The usual proof is a Baire category argument. Let X and Y be complete metric spaces and f_n : X --> Y continuous, with f_n --> f pointwise. Put P_m(e) = {x in X : |f_m(x) - f(x)| <= e}, G(e) = union of INTERIOR OF P_m(e) for m=1 to infinity. The idea is to show that the set of x at which f is continuous is intersection of G(e) for e > 0, which is the same as intersection of G(1/n) for n = 1, 2, .... Next define F_m(e) = {x : |f_m(x) - f_n(x)| <= e for all n >= m}. Then F_m(e) is closed and F_m(e) \subset P_m(e), hence int F_m(e) \subset int P_m(e). Furthermore, since f_n(x) --> f(x) for each x, we have X = union F_m(e), m = 1 to infinity. The sets F_m(e) \ int F_m(e) are nowhere dense (they're closed, and have empty interior), thus W(e) = union of F_m(e) \ int F_m(e), m = 1 to infinity is of the first category in X, hence W(e) \supset X \ union of int F_m(e) \supset X \ union of int P_m(e) i.e. W(e) \supset X \ G(e). Now take the union of the W(e) for e = 1, 1/2, 1/3, ... to conclude X - intersection of G(1/n) is of first category, i.e. the set of points of discontinuity is of the first category. This has all kinds of neat consequences. For example, if T_n is a continuous linear operator from one Banach space X to another, and T_n converges pointwise to T, then T must be continuous at SOME point, hence at ALL points; this is the most-frequently-used consequence of the Uniform Boundedness Principle. (Of course, it's not the most general phrasing of UBP.) Also, the Vitali-Hahn-Saks theorem follows easily: if mu_n is a finite (positive) measure on a measurable space (X,Sigma) such that mu_n --> mu pointwise in R (so in particular mu is real-valued) then mu is a measure. Build a measure nu(E) = \sum_n mu(E)/(mu_n(X)*2^n), note that each mu_n is absolutely continuous with respect to nu; note that the space of sets in Sigma with nu(E) < infinity (i.e. all of them) is a pseudometric space with the pseudometric d(A,B) = nu((A\B) U (B\A)), and is complete; and the absolute continuity of the mu_n w.r.t. nu implies their continuity on this metric space; hence the limit, mu, is continuous at SOME point; continuity at SOME point implies continuity at the empty set; which in turn implies continuity at all points; which in turn implies mu is a measure. A really, really neat proof. --Ron Bruck -- Due to University fiscal constraints, .sigs may not be exceed one line.