Hi again,
I did not understand your hint. It sounds like it applies to (a), but
I find it much easier to prove (a) without using a "heavy" theorem
like the dominated convergence theorem:
(a) I use the triangle inequality for integrals, which says that
integral(f) < integral(f)
(here and below, for convenience I use "<" to mean "< or =")
(If you don't know the proof of this, tell me and I'll write it for
you.) Anyway, we need to show that integral(fn)>integral(f), or
integral(fn)integral(f)>0. But
integral(fn)integral(f) = integral(fnf) < integral(fnf) <
< mu(X)*sup(fn(x)f(x))
where the supremum is over all x in X.
fnf converges uniformly, so for any e > 0 (e=epsilon) there is an N
such that for all n>N, we have fn(x)f(x)<e for any x in X, in other
words
supfn(x)f(x) < e. But then for all n>N
integral(fn)integral(f) < mu(X)*e
Since e>0 is arbitrary this proves that integral(fn)>integral(f).
(here I am using the fact that mu(X)<infinity, otherwise the above
inequality says nothing.)
(b) Take (X,M,mu)=(R=reals, B=borel sets, mu=Lebesgue measure). Take
fn = constant function 1/n. Then fn>0 uniformly, but
integral(fn) = infinity /> integral(0)=0.
(c) Take (X,M,mu)=([0,1],B=borel sets,mu=Lebesgue measure on [0,1]).
Take
fn = n*indicator function of (0,1/n). Then fn>0 pointwise for any x
(for x=0, fn(x)=0 for all n; for x>0 fn(x)=0 for any n>1/x), but
integral(fn) = 1 (for all n) /> integral(0)=0
Regards,
dannidin 
Clarification of Answer by
dannidinga
on
09 Dec 2002 00:23 PST
Sure I can:
f is a bounded function, let's say f<M. fn converges to f uniformly,
so there exists an N such that for all n>N we have fnf < 1 for all
x in X. Therefore, for all n>N we have fn< M + 1 by the triangle
inequality. So the sequence fn is uniformly bounded (you can have it
uniformly bounded even for all values of n by replacing M+1 by a
bigger constant that also bounds f1, f2, ..., fN). Now we can apply
the dominated/bounded convergence theorem to say that
integral(fn)>integral(lim(fn))=integral(f) as n>infinity.
Note that we can only use the dominated convergence theorem because
mu(X)<infinity, otherwise the constant function g=M+1 that dominates
the sequence is not an integrable function.
Regards,
dannidin
