Postby **blp** » Mon Oct 27, 2008 7:28 am

There's a pretty typical counterexample for the second problem, though writing down the specific equation is a bit harder.

So think about a function that is otherwise constantly zero, but it contains a thin very high smooth "spike" which has area one. We can assume that for f_n the spike is e.g. centered around (n-1)/n and has width 1/n^2 with height large enough for the integral over it to be 1 (draw a picture).

Now if we take any point x in R. If x>=1, then for all n f_n(x)=0 and for all x<1 the spike has gone "past" x for large enough n, so that there's a k>0 with f_n(x)=0 if n>k. This means that f_n->0 pointwise. But we have

int_0^1 f_n = 1 for all n

It means that F_n(x)=1 for all x>=1 while F(x)=0 for all x. What this means is that (a) and (c) are incorrect, but (b) is correct because F'_n=f_n which converges to f=0 pointwise. Thus assuming that the limit function is continuous doesn't make the problem behave any better, so we really need uniform convergence.