Re: L^1 function question...

On Mon, 9 May 2005 10:17:35 -0500, "Jon Slaughter"
<Jon_Slaughter@xxxxxxxxxxx> wrote:

>
>"David C. Ullrich" <ullrich@xxxxxxxxxxxxxxxx> wrote in message
>news:earu71prju31mha6mn08v6of1163vehqq6@xxxxxxxxxx
>> On Mon, 9 May 2005 08:21:25 -0500, "Jon Slaughter"
>> <Jon_Slaughter@xxxxxxxxxxx> wrote:
>>
>>>
>>>"James" <James545@xxxxxxxxx> wrote in message
>>>news:d5nnco$5nnq$1@xxxxxxxxxxxxxxxxxxxx
>>>> Let f_n be a sequence of integrable real-valued functions on (0,1).
>>>> Suppose that
>>>>
>>>> (a) f_n(x) <= x^(-1/2), 0 < x < 1, n = 1, 2, 3, ...
>>>> (b) f_n ----> f a.e. on (0,1)
>>>> (c) lim integral_(0,1) f_n(x) dx exists and is finite.
>>>>
>>>> Prove that f is integrable on (0,1).
>>>>
>>>> By using Fatou on the family x^(-1/2) - f_n(x), I found that
>>>> integral_(0,1) f(x)dx >= lim integral_(0,1) f_n(x)dx
>>>>
>>>> But how can I show that f is integrable on (0,1)?
>>>
>>>Because the difference between f and f_n is at most a set of measure 0
>>>hence
>>>the integrals of f and f_n are the same (in the limit ofcourse).
>>
>> This doesn't make any sense.
>>
>
>
>
>By LDCT, int(lim f_k) = lim int(f_k)
>but since lim f_k = f a.e., and the integral of any null set is 0,
>
>int(lim f_k) = lim(f + S) = int(f)
>
>hence lim int(f_k) = int(f) and since lim int(f_k) is integrable, so is
>int(f).
>
>Actualy, it just follows from the generalized LDCT.

_That_ makes sense. It's wrong, as Jose has been trying to
explain, but it makes sense. (Except for the question of
what "S" is...)

Doesn't change the fact that what you said previously made
no sense - if "Because the difference between f and f_n is at
most a set of measure 0 hence the integrals of f and f_n
are the same (in the limit ofcourse)" we could for example
prove that int f_n -> int f whenever f_n -> f ae.

>>>> Thanks!
>>>>
>>>> James
>>>>
>>>
>>
>>
>> ************************
>>
>> David C. Ullrich
>


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

David C. Ullrich
.

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