Re: lebesgue
- From: v_aylin2000@xxxxxxxxxxx
- Date: Thu, 21 Feb 2008 10:14:48 -0800 (PST)
On 21 Şubat, 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
.
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