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!

  • $\begingroup$ You have $f_n(0) = 0$ for all $n$, so you'd rather get $$\begin{cases} 0 & x = 0\\ \infty & 0 < x \leqslant 0\\ 0 & 0 < x \leqslant 1\end{cases}$$ But really, fix one (arbitrary) point $\xi$ and look at the sequence $f_n(\xi)$. It's pointwise convergence after all. $\endgroup$ – Daniel Fischer Sep 28 '13 at 22:23

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.

  • $\begingroup$ I can see the motivation for $x>1/n$. My confusion is with anything below $1/n$. Also do all such functions form a triangle? Thank you. $\endgroup$ – RDizzl3 Sep 28 '13 at 22:32
  • $\begingroup$ With $x < 1/n$, $f(x)$ can indeed be quite large (arbitrarily so, depending on $n$, but still for any particular $n$ finite). About the triangles: yes, it might be helpful to draw a picture. $\endgroup$ – AlexM Sep 28 '13 at 22:41

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