Can You Solve the Non-Uniform Convergence Integral Challenge?

[Ackbach]

Here is this week's POTW:

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Find a sequence of functions $\{f_n\}$ on $[0,1]$ such that
$$\lim_{n\to \infty}\int_0^1 f_n(x) \, dx=\int_0^1 \lim_{n\to \infty} \, f_n(x) \, dx,$$
but $\{f_n\}$ does not converge uniformly to any function $f(x)$ on $[0,1]$. Thus, uniform convergence is not a necessary condition for convergence of the integrals.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

Congratulations to johng for his correct solution, which is below:

Spoiler

For each positive integer $n$, let $f_n(x)$ be $0$ on $[0,1-1/n]$, $f_n(1)=1$ and $f$ is linear on $[1-1/n,1]$. Then the pointwise limit of $f_n$ is the function $f$ where $f$ is $0$ on $[0,1)$ and $f(1)=1$. Since each $f_n$ is continuous and $f$ is not continuous, the convergence is not uniform. Now
$$\int_0^1 f_n(x)\,dx={1\over 2n} \text{ and }\int_0^1 f(x)\,dx =0$$
So the limit of the integrals is the integral of the limit. QED.