hi dannidin-ga could you please help me in this problem:


     give an example showing that "convergence in measure' of a
sequence {fn}running from n=1 to infinity subset of M does not imply
convergence of {fn(X)} running from n=1 to infinity at any point x
belongs to X.

  So, in particular convergence in measure does not imply convergence
almost everywhere.
           also give an example of a sequence {fn} running from n=1 to
infinity
subset of L'  which converges in measure to a function f belongs to L'
but such that not even a subsequence of {fn}running from n=1 to
infinity is convergent in L'.

Hi Madukar-ga,

For your first question, believe it or not but I already answered it!
Check out

https://answers.google.com/answers/main?cmd=threadview&id=120524

The sequence of functions that I construct there (in part b.)
converges in measure to 0, since the set of points where the function
fn is not zero has measure 1/2^k, where k is approximately log(n) (to
base 2). However, for EVERY point x in [0,1], fn(x) is not a
convergent sequence of real numbers.

Now for the second question: Define the sequence of functions
gn:[0,1]->R by

gn(x) = 0    if x > 1/2^n

      = 2^n  if x < or = 1/2^n

Then gn converges in measure to 0 (for the same reason as stated above
for fn). But gn cannot converge in L' to 0, and not even any
subsequence can converge to 0 in L', because for all n we have !!gn!!1
= !!gn - 0!!1 = 1
(!!gn - 0!! measures the distance in L' between gn and 0. If this does
not converge to zero then gn does not converge in L' to 0)

Hope this is clear, if not please ask and I'll clarify.
Regards,
dannidin