Hi Madukarga,
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
nonnegative "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)<infinity
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.
dannidin 
Clarification of Answer by
dannidinga
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.)
dannidin
