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)| = |integral(fn-f)| < integral(|fn-f|) <
where the supremum is over all x in X.
|fn-f| 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
sup|fn(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]).
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
Clarification of Answer by
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 |fn-f| < 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.