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 Subject: INTEGRATION OF NON-NEGATIVE FUNCTIONS 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.```
 ```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 zero. 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 integral(g_n)<=integral(f)
 madukar-ga rated this answer: `I really appreicate for your kind help.`