For $n\in{\mathbb{N}}$ let $$f_n(x)=nx(1-x^2)^n\qquad(0\le x\le 1).$$
Show that $\{f_n\}_{n=1}^\infty$ converges pointwise to $0$ on $[0,1]$.
Show that $\{\int_0^1f_n\}_{n=1}^\infty$ converges to $\frac12$.
I've already shown both of these statements to be true. What I don't understand is this: how can $f_n$ converge pointwise to $0$, yet the sequence $\{\int_0^1f_n\}_{n=1}^\infty$ converges to $\frac12$? Isn't that almost like saying $f(x)=0\qquad(a\le x\le b)$, but $\int_a^bf(x)=\frac12$?
Clearly that would be false. I know this has something to do with that this is a sequence of functions but it still baffles my mind. Thanks in advance.
Best Answer
Pointwise convergence of a sequence of functions doesn't necessarily imply convergence of their integrals.
Here's a simpler example. Let $f_n$ be a rectangle over the interval $(0,1/n]$ with height $n$. Then the $f_n$ get narrower and taller as $n\to\infty$, maintaining an area of $1$ for every $n$. So $\int_0^1f_n\to1$ as $n\to\infty$. But you can see that the $f_n$ converge pointwise to zero everywhere, since for every $x>0$ there will be a point past which all the rectangles are so narrow that they've already 'squeezed past' $x$. These $f_n$ have the odd property that all their area 'escapes' as they converge, so that their limit has area zero.
BTW, there is a condition on the $\{f_n\}$ that assures that pointwise convergence of $f_n$ to $f$ implies $\int f_n\to\int f$: it's called 'uniform integrability'.