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Q: INTEGRATION OF NON-NEGATIVE FUNCTIONS ( Answered 5 out of 5 stars,   0 Comments )
Category: Science > Math
Asked by: madukar-ga
List Price: $3.00
Posted: 14 Nov 2002 12:47 PST
Expires: 14 Dec 2002 12:47 PST
Question ID: 107844
If f belongs to L+(L+ = the space of all measurable functions from X
to [0,infinity], and integral f < infinity,then {x: f(x)= infinity}is
a null set and {x:f(x)>0}is sigma finite.
Answered By: dannidin-ga on 15 Nov 2002 01:05 PST
Rated:5 out of 5 stars
Hi Madukar-ga,

I gather from the question that X is a measure space, and denote by
m(.) the accompanying measure.

Claim 1. {x: f(x) = infinity} (call this set A) is a set of measure

Proof: Recall that, for a function f in L+, the integral of f is
defined as the supremum of the integral of g, as g goes through all
non-negative "simple" functions which are <= (less than or equal to)
f. A simple function is a function which can be written as sum(a_i
1(A_i)) where a_i are real numbers and 1(A_i) is the indicator
function (sometimes called characteristic function) of a measurable
set A_i. For a simple function g=sum(a_i 1(A_i)), its integral is
defined as sum(a_i m(A_i)).

Assume by contradiction that m(A)>0. Denote I = integral(f) <
infinity. Then the function g = 2*I/m(a) * 1(A) is a simple function,
which is <= f (since f is infinity on A). Therefore
integral(g)<=integral(f). But

integral(g) = 2*I/m(a) * m(A) = 2*I, whereas
integral(f) = I < integral(g). We have a contradiction, therefore
m(A)=0. QED

Claim 2. {x:f(x)>0} (call this set B) is a countable union of sets of
finite measure.

Proof: Define B_n = { x: f(x)>1/n }  (n= 1,2, ...). Clearly B is the
(countable) union of B_n. But for each n=1,2,..., the function

g_n = (1/n) * 1(B_n)

is a simple function which is <= f, therefore 


But integral(g_n) = (1/n) * m(B_n)

So m(B_n) < infinity, and we have exhibited B as a countable union of
sets of finite measure. QED

Again, if there is anything that is unclear do not hesitate to ask for
more details.


Request for Answer Clarification by madukar-ga on 15 Nov 2002 21:13 PST

  I didnot understand what does the sysmbol 2*I/m(a)*m(a) stands for?
is it 2 to the power of I/m(a) OR 2 times I/m(a). and also what does
a_i means? is it  ai,a2.... like that?  and also what does QED stands

  I would be thankful to you if you clafiy these minor doubts.

Clarification of Answer by dannidin-ga on 16 Nov 2002 01:11 PST
Hi Madukar,

Remember, I denoted by "I" the (numerical) value of the integral of f.

1. "2*I/m(A) * m(A)" is "two times I over m(A) times m(A)" - sorry I
wrote "a" instead of capital A.

2. a_i means "a subscript i" or "a sub i". It is a notation for a
generic element in a sequence: If a_1, a_2, a_3, ... (read: "a sub 1,
a sub 2, ...") are the elements of a sequence, then a_i would be the
"i'th" element of the sequence. i is an "index" that goes over the
values 1,2,3, ...

3. QED is the traditional mathematical symbol for "end of proof". It
is shorthand for the latin expression "quod erat demonstrandum", which
means "that which was to be demonstrated".

Hope this makes it clearer, (please ask for clarification again if you
still have doubts.)
madukar-ga rated this answer:5 out of 5 stars
I really appreicate for your kind help.

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