Re: Can't find the difference between 'Converge almost uniformly' and 'Uniform convergent almost everywhere'.

On 6 Aug 2005 11:32:01 -0700, "mathlover" <2000korea@xxxxxxxxxxx>
wrote:

>
>> Here is an example. E = (0,1), f_n(x) = 1/(nx), f(x) = 0.
>> f_n converges to f uniformly on any set of the form (epsilon,1).
>> But there is no set A of measure 1 such that f_n converges
>> uniformly to f on A.
>>
>
>Thank you so much for the helpful example. I think I learned the
>concept.
>The following is what I understood.
>
>If F_n is 'uniformly convergent a.e' to F on given set E then
>F_n is converging uniformly on A\E but the measure of set A is zero.

You meant E\A, not A\E.

>If F_n is 'almost uniformly convergent' to F on given set E then
>F_n is converging uniformly on A\E but the measure of set A is less
>than
>epsilon.

You should get in the habit of not leaving out the little
words - they're very important, leaving them out leads to
the sort of confusion in your original post. You meant that
for every epsilon > 0 there exists A such that the measure
of A is less than epsilon and F_n converges uniformly on
E\A.

>By definition It's trivial to see 'uniformly convergent a.e' implies
>'almost uniform convergent'.
>For the converse we can control the 'epsilon' as small as possible.
>So the limit of the measure (non uniform converging part) tends to
>zero.
>'almost uniform convergent' implies also 'uniformly convergent a.e'?

No. The converse is _false_. Edgar _gave_ a counterexample!

************************

David C. Ullrich
.

Relevant Pages