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Q: integrationof complex functions ( Answered 4 out of 5 stars,   0 Comments )
Subject: integrationof complex functions
Category: Science > Math
Asked by: madukar-ga
List Price: $10.00
Posted: 06 Dec 2002 14:21 PST
Expires: 05 Jan 2003 14:21 PST
Question ID: 120524
let (fn)where n belongs to N be a sequence of Lebesgue measurable
   fn : [0,1]to [0,1].

(a) prove that if fn tends to 0 a.e, then integral fn over [0,1] tends
to 0.

(b) show that the reverse implicatoin is false by constructing a
sequence of Lebesgue meaurable functions fn :[0,1] to [0,1] with
integral fn over [0,1] such that (fn)where n belongs to N does not
converge at any point x belongs to [0,1].
Subject: Re: integrationof complex functions
Answered By: dannidin-ga on 07 Dec 2002 01:07 PST
Rated:4 out of 5 stars
Hi Madukar-ga,

(a) fn is a sequence of bounded measurable functions that converges to
0 a.e., therefore by the bounded convergence theorem

 lim (integral fn) = integral(lim fn) = integral(0) = 0

(b) Define a sequence of indicator functions of intervals as follows:

I[a,b] denotes the indicator/characteristic function of the interval
[a,b], i.e. the function that is equal to 1 on [a,b] and 0 elsewhere.

  f0 = I[0,1]

  f1 = I[0,1/2]
  f2 = I[1/2,1]
  f3 = I[0,1/3]
  f4 = I[1/3,2/3]
  f5 = I[2/3,1]

  f6 = I[0,1/4]
  f7 = I[1/4,1/2]
  f8 = I[1/2,3/4]
  f9 = I[3/4,1]

  f10 = I[0,1/5]
  f11 = I[1/5,2/5] ... 

  fn = I[k/m,(k+1)/m] if n = m(m-1)/2 + k

Then integral(fn) = 1/m (for n=m(m-1)/2+k) ---> 0 as n->infinity.
However, clearly for any fixed x in [0,1], the sequence of number
fn(x) cannot converge, because there are infinitely many values of n
for which fn(x) is zero (any n such that x is not contained in the
interval [k/m,(k+1)/m]) and infinitely many values of n for which
fn(x) is 1 (any n such that x IS in the abovementioned interval).

(A professor of mine used to call this example "the Goodyear blimp",
because the graphs of the function sequence fn look like a blimp that
passes back and forth in the sky...)

Hope this helps, thanks for the question. I'm here if you need any

madukar-ga rated this answer:4 out of 5 stars
good it really helped me a lot

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