TITLE: Pointwise convergence to zero, with integrals converging to a nonzero value
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: Perhaps easier to visualize: Let $f_n$ be a triangular spike over $[1/(2n),1/n]$ of height $2n.$ Fixing any $x\in (0,1],$ these triangles eventually all lie to the left of $x.$ Hence $f_n(x)$ is eventually $0.$ Also, $f_n(0) = 0$ for every $n.$ Thus $f_n \to 0$ pointwise on $[0,1].$ But $\int_0^1f_n = 1/2$ for every $n.$