From: Ronald Bruck <bruck@math.usc.edu>
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 <michaelknudsen@city.dk> 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

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