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
clarifications.
Cheers,
dannidin |