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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.)

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  • $\begingroup$ 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 $\endgroup$
    – b00n heT
    Commented Dec 15, 2016 at 6:14
  • $\begingroup$ @b00nheT: The convergence is not uniform, otherwise it would be straightforward. $\endgroup$
    – copper.hat
    Commented Dec 15, 2016 at 6:16
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    $\begingroup$ 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. $\endgroup$
    – Robert Israel
    Commented Dec 15, 2016 at 7:08
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    $\begingroup$ For future people looking for @RobertIsrael link but seeing that it is paid. Here is a link to the pdf $\endgroup$
    – user160110
    Commented Dec 15, 2016 at 7:12
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    $\begingroup$ Osgood also proved the continuous case, as Luxemburg remarks. Here is a link to Osgood's paper. $\endgroup$
    – Robert Israel
    Commented Dec 15, 2016 at 7:25

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