Let $f_n\in C([0,1])$ be a sequence of functions converging uniformly to a function $f$. Show that
$$\lim_{n\rightarrow\infty}\int_0^1f_n(x)dx = \int_0^1 f(x)dx.$$
Give a counterexample to show that the pointwise convergence of continuous functions $f_n$ to a continuous function $f$ does not imply the convergence of the corresponding integrals. I have the following counterexample:
$$f_n(x) = \begin{cases} 2n^2 x & 0\leq x \leq\frac{1}{2n} \\
-2n^2(x-\frac{1}{n}) & \frac{1}{2n}\leq x \leq \frac{1}{n} \\
0 & \frac{1}{n}\leq x \leq 1 \end{cases}$$
I am having trouble seeing how this function converges to 0. I may be looking at this wrong but when you take $n\rightarrow\infty$ doesn't this function look like this?
$$f_n(x) = \begin{cases} \infty & 0\leq x \leq 0 \\
\infty & 0\leq x \leq 0 \\
0 & 0 \leq x \leq 1 \end{cases}.$$
as $n\rightarrow\infty$. Where I am having trouble seeing $f_n\rightarrow 0$. Am I looking at the convergence of functions in a wrong way? If so how should I consider seeing how such functions converge?
I know the integral of $\int_0^1 f_n = \frac{1}{2}$ and $\int_0^1\lim_{n\rightarrow\infty}f_n = 0$. Also I don't need help with the proof concerning integrals. Just on the counterexample.
Thank you for any help and comments!
Best Answer
Take a point $x \in [0,1]$. Suppose $x > 1/n$. Then $f_n(x) = 0$ because $f_n(z) = 0$ for $z > 1/n$. Note that this convergence isn't uniform, since $x$ can be quiite close to zero.
So the sequence of functions converges to zero, and the integral of zero is zero. But the integral of any one of $f_n$ is actually (using the area under a triangle is $1/2ab$) turns out to be 1/2, so the limit of integrals of our sequence does get near zero.