# Proving $\lim\limits_{n\to \infty}\int\limits_{0}^{1} f_n(x)dx=0$

If $$\{f_n\}$$ is a sequence of continuous functions on $$[0,1]$$ such that $$0\leq f_n\leq 1$$ and such that $$f_n(x)\to 0$$ as $$n\to \infty$$, for every $$x\in [0,1]$$, then $$\lim\limits_{n\to \infty}\int\limits_{0}^{1} f_n(x)dx=0$$

Try to prove this without using any measure theory or any theorems about Lebesgue integration.

I am quite stuck as to what method I should use to go about solving this proof. I am thinking of showing that the set $$A_n=\{x\in[0,1]: f_n(x)\geq \epsilon/2\}$$ has a measure smaller than $$\epsilon/2$$ for large n. But proving that especially without measure theory is pretty hard. Anyone have any hints?

(I would like to thank everyone that offered help. You were all fantastic.)

• No measure theory is needed: for any $\epsilon>0$ find $n$ such that $f_n(x)<\epsilon$ (Use compactness). Then use the classical maximum times length argument to get an upper bound Commented Dec 15, 2016 at 6:14
• @b00nheT: The convergence is not uniform, otherwise it would be straightforward. Commented Dec 15, 2016 at 6:16
• There is Arzela's Dominated Convergence Theorem for the Riemann integral. I suspect this exercise is meant to give you some appreciation for how much nicer the Lebesgue integral is than the Riemann integral from this point of view. Commented Dec 15, 2016 at 7:08
• For future people looking for @RobertIsrael link but seeing that it is paid. Here is a link to the pdf Commented Dec 15, 2016 at 7:12
• Osgood also proved the continuous case, as Luxemburg remarks. Here is a link to Osgood's paper. Commented Dec 15, 2016 at 7:25