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Q: measure theory ( Answered 5 out of 5 stars,   0 Comments )
Subject: measure theory
Category: Science > Math
Asked by: madukar-ga
List Price: $10.00
Posted: 06 Dec 2002 14:11 PST
Expires: 05 Jan 2003 14:11 PST
Question ID: 120521
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
 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.
Subject: Re: measure theory
Answered By: dannidin-ga on 07 Dec 2002 01:24 PST
Rated:5 out of 5 stars
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
   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


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

    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.

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.

madukar-ga rated this answer:5 out of 5 stars
outstanding answer.

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