Re: lebesgue

On Feb 21, 2:28 pm, v_aylin2...@xxxxxxxxxxx wrote:
On 21 Şubat, 14:27, The World Wide Wade <aderamey.a...@xxxxxxxxxxx>
wrote:



In article
<432dbdab-0a93-4a80-9764-b47235105...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

v_aylin2...@xxxxxxxxxxx wrote:
On 21 Șubat, 13:06, Robert Israel
<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
v_aylin2...@xxxxxxxxxxx writes:
f is a nonnegative and integrable over a measurable set E. Then
epsilon greater than 0 and delta greater than 0 s.t. over a measurable
set A subset of E with
lamda(A)<delta, we have integral_A f d lambda < epsilon

it means A has measure zero.If A has measure zero we can say lambdaA=0
then,
lambdaA=0 and f is measurable integral_R f X_A dlambda=0

we know that if A subset of E
integral_A f dlambda= integral_R X_A f dlambda<=integral_R X_E
dlambda=
integral_E f dlambda

from this
0<= integral_R X_E dlambda=integral_E f dlambda

What should I do now?

That depends on what you're trying to do, which is not at all clear.
What exactly is the problem you're trying to solve? Please be careful
to include the quantifiers (for all) and (there exists), and in the correct
order: they are very important.
--
Robert Israel isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada- Alıntıyı
gizle -

- Alıntıyı göster -

I'm trying to show that

∫f dlambda<epsilon
A

State the problem precisely.- Alıntıyı gizle -

- Alıntıyı göster -

f is a nonnegative and integrable over a measurable set E. Then є>0
and δ>0 s.t. over a measurable
set A subset of E with λ(A)<δ, show that ∫fdλ<є
A

You didn't take my advice, nor did you take Wade's. There may
be a language barrier here, but I suspect it shows a basic lack
of understanding of the mathematics.

Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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