From: "David C. Ullrich" <ullrich@math.okstate.edu>
Subject: Re: Convergence in measure
Date: Tue, 07 Mar 2000 16:01:33 -0600
Newsgroups: sci.math
Summary: [missing]

    It really is necessary for the measure space to be finite.
Actually it's not necessary for one of the conclusions but
it is for the other.
    When you see the proofs you will see where the finite
measure comes in, which might be a hint what a counterexample
might look like. Hints:

|(fn + gn) - (f + g)| <= |fn - f| + |gn - g|

is what you need for one of the results. For the
other it's the same trick as always when fn is close
to f and gn is close to g, and you want to conclude
fn*gn is close to f*g:

|(fn*gn) - (f*g)| = |fn*gn - fn*g + fn*g - f*g|
                        <= |fn|*|gn - g| + |g|*|fn - f|


jonas24y@my-deja.com wrote:

> Hello there!
>
> I wonder how to prove the following:
>
> In a finite measure space, let {fn} and {gn} be sequences of almost
> everywhere real-valued, measurable functions that converge in measure to
> f and g, respectively. If a and b are any real numbers, then {afn + bgn}
> converges in measure to af + bg, and {fng} converges in measure to fg.
>
> Also, is it really necessary for the space to be finite in both cases,
> or can it in fact be infinite?
>
> Thanks for any replies or hints!
>
> Sent via Deja.com http://www.deja.com/
> Before you buy.
==============================================================================

From: "David C. Ullrich" <ullrich@math.okstate.edu>
Subject: Re: Convergence in measure
Date: Fri, 17 Mar 2000 13:10:53 -0600
Newsgroups: sci.math

jonas24y@my-deja.com wrote:

> Hello! Thanks for the answer, but I am still a bit "clueless"
>
> In article <38C57C3D.4AC4E84E@math.okstate.edu>,
> "David C. Ullrich" <ullrich@math.okstate.edu> wrote:
> > It really is necessary for the measure space to be finite.
> > Actually it's not necessary for one of the conclusions but
> > it is for the other.
> > When you see the proofs you will see where the finite
> > measure comes in, which might be a hint what a counterexample
> > might look like. Hints:
> >
> > |(fn + gn) - (f + g)| <= |fn - f| + |gn - g|
> >
> > is what you need for one of the results. For the
> > other it's the same trick as always when fn is close
> > to f and gn is close to g, and you want to conclude
> > fn*gn is close to f*g:
> >
> > |(fn*gn) - (f*g)| = |fn*gn - fn*g + fn*g - f*g|
> > <= |fn|*|gn - g| + |g|*|fn - f|
>
> I understand that the finiteness of the space comes in here, but I dont
> see HOW. Can you please explain some more?

    Suppose that epsilon > 0, and you want to show that
|(fn*gn) - (f*g)| < epsilon. (At a point, or on a large set,
or whatever, depending on what point in the argument
we're at).  The standard way to do that would be
to use the above, and say you only need to show that

(*) |fn|*|gn - g| + |g|*|fn - f| < epsilon.

Now what would you need to know about f, g, fn, and gn
to get (*)? All you need to know is that |gn - g| and
|fn - f| are both small? Yes and no: _how_ small do they
have to be to get (*) ???

[Quote of original article deleted -- djr]