From: "G. A. Edgar" <edgar@math.ohio-state.edu.nospam>
Subject: Re: signed measures
Date: Wed, 12 May 1999 10:54:18 -0400
Newsgroups: sci.math.research
Keywords: convergence of signed measures does not imply convergence of parts

In article <3738DC78.B3E7BB85@asuvm.inre.asu.edu>, Hector Chade
<Hector.Chade@asu.edu> wrote:

> I have the following question:
> 
> "Let $(X,B)$ be a measurable space and let $\{\nu_{n}\}$ be a sequence
> of finite signed measures converging setwise (i.e., for each measurable
> set) to a signed measure $\nu$. Let $\nu_{n}^{+}-\nu_{n}^{-}$  and
> $\nu^{+}-\nu^{-}$ be the Jordan decomposition of $\nu_{n}$ and $\nu$,
> respectively.
> Does the positive (negative) variation of $\nu_{n}$ converge setwise to
> the positive (negative) variation of $\nu$ (i.e., is it true that
> $\nu_{n}^{+}$ converges to $\nu^{+}$ and $\nu_{n}^{-}$ converges to
> $\nu^{-}$ as well?)?"
> 
> Thankfully,
> 
> Hector

Let's try X=[0,1], B=Borel sets, and nu_n defined by the Haar
functions:  d nu_n(x) = h_n(x) dx.  Haar functions look like
h_n(x) = sign(sin(2 pi 2^n x)).  Now doesn't nu_n converge
setwise to zero?  (Think of h_n as an orthonormal sequence.)
Certainly the positive and negative parts do not converge to zero.

-- 
Gerald A. Edgar              edgar@math.ohio-state.edu
Department of Mathematics    telephone: 614-292-0395 (Office)
The Ohio State University      614-292-4975 (Math. Dept.)
Columbus, OH 43210             614-292-1479 (Dept. Fax)
==============================================================================

From: "David C. Ullrich" <ullrich@math.okstate.edu>
Subject: Re: signed measures
Date: Thu, 13 May 1999 18:16:47 GMT
Newsgroups: sci.math.research

In article <120519991054186253%edgar@math.ohio-state.edu.nospam>,
  "G. A. Edgar" <edgar@math.ohio-state.edu.nospam> wrote:
[previous article quoted in full -- djr]

      Not critically important, but for the benefit of the
students out there, I believe these would more properly be
called the Rademacher functions; the Haar functions are
something else.

      (A Haar function is supported on a dyadic interval, not
on all of [0,1]. They form an orthonormal basis for L2([0,1]).
(Which is how I remember which is which - people talk about the
Haar basis, they don't talk about the Rademacher basis...))


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