let (X,M,mu)be a measure space with mu(X) < infinity, (fn) n belongs
to N a sequence of bounded measurable functions fn : X to R, f :X to R
with
 fn converges to f uniformly on X.

  (a) show that integral fn converges to integral f.

  (b) construct a counterexample showing that the assumption mu (X) <
infinity cannot be dropped.

  (c) construct a counterexample showing that the assumption of
uniform convergence cannot be relaxed to pointwise convergence.


  hint: verify that the function fn are unifromly bounded,then apply
the dominated onvergence theorm.

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(fn-f)| < integral(|fn-f|) < 
         < mu(X)*sup(|fn(x)-f(x)|)

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
words
   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]).
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

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Request for Answer Clarification by madukar-ga on 08 Dec 2002 14:00 PST

hi,

    could you please  prove the theorm by verify that the given
function fn is uniformly bounded and by dominated convergence theorm.

   waiting for your reply.

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Clarification of Answer by dannidin-ga 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 |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.

Regards,
dannidin

outstanding answer.