If f belongs to L+(the space of all measurable functions from X to
[0,infinity] ) and integral f < infinity, for every eplison >0 there
exists E belongs to Script M such that mu(E)< infinity and integralto
the base E f >(integral f) - epsilon.

Hi again Madukar-ga,

For n=1,2,3,..., define the set E_n = {x in X : f(x) < n}

Define the function g_n = f * indicator (=characteristic) function of
E_n.

g_n is a sequence of positive functions that converges pointwise to f,
since for any x in X, if f(x)<n_0 then g_n(x)=f(x) for all n>n_0. Also
all the g_n are bounded by f, which is an integrable function.
Therefore by the dominated convergence theorem, we know that
integral(g_n)-->integral(f) as n->infinity.

In particular, for some value of n we will have that

integral(g_n) > integral(f)-epsilon.

But integral(g_n) = integral(f) over the set E_n. So E_n is the E that
you asked for.

Hope everything's clear, if not I'm here for you...

Regards,
dannidin