Prove that $m(U(n,\epsilon)) \xrightarrow{n\to\infty} 0$ where $U(n,\epsilon) = \{x: f_n(x) > \epsilon\}$ 
Suppose $(f_n)_{n\in\mathbb N}$ is a sequence of continuous functions on $[0,1]$ such that

*

*$f_n(x) \xrightarrow{n\to\infty} 0$ for every $0\le x\le 1$,

*$0\le f_n\le 1$ for every $n\in\mathbb N$.

Define $U(n,\epsilon) = \{x: f_n(x) > \epsilon\}$ for $\epsilon > 0$. Prove that $$m(U(n,\epsilon)) \xrightarrow{n\to\infty} 0$$ where $m$ is the restriction of the Lebesgue measure to $[0,1]$.

Try not to use anything fancy such as the Dominated Convergence Theorem (DCT). Try to work with the assumptions directly, such as continuity of $f_n$ and compactness of $[0,1]$.

I didn't make substantial progress, but let me collect all my thoughts here. I saw that $$\bigcap_{n=1}^\infty U(n,\epsilon) = \varnothing$$
If we could use the DCT, I would do (hope it's correct) $$\lim_{n\to\infty} m(U(n,\epsilon)) = \lim_{n\to\infty}\int_0^1 \mathbf{1}_{U(n,\epsilon)} \mathrm{d}m = \int_0^1 \lim_{n\to\infty}\mathbf{1}_{U(n,\epsilon)} \mathrm{d}m = 0$$
where $\mathbf{1}_A$ is the indicator function of set $A$. However, I'm looking for ways to prove this without DCT. Another thing I noticed is that $(U(n,\epsilon))^c = [0,1]\setminus U(n,\epsilon)$ is compact, since it is closed and bounded. It is obviously bounded since it is contained in $[0,1]$ and it is closed since it is the continuous pull-back of a closed set, namely $[0,\epsilon]$. Clearly, $U(n,\epsilon)$ is open.
Any thoughts on what to do next? Thank you.
 A: This solution still uses "fanciful" properties of measures, that is $A_n\searrow A$ and $\mu(A_1)<\infty$, then $\mu(A)=\lim_n\mu(A_n)$.
The set where $f_n\xrightarrow{n\rightarrow\infty}0$ can be expressed as
$$G=\bigcap^\infty_{k=1}\bigcup^\infty_{\ell=1}\bigcap^\infty_{j\geq \ell}\Big\{|f_j|\leq\frac{1}{k}\Big\}$$
By assumption $G=[0,1]$.
Then
$$U=[0,1]\setminus G=\bigcup^\infty_{k=1}\bigcap^\infty_{\ell=1}\bigcup^\infty_{j\geq \ell}\Big\{|f_j|>\frac{1}{k}\Big\}=\emptyset$$
Consequently, for any $k$,
$$m\big(\bigcap^\infty_{\ell=1}\bigcup^\infty_{j\geq \ell}\Big\{|f_j|>\frac{1}{k}\Big\}\Big)=0$$
Since $m$ is finite
$$0=m\big(\bigcap^\infty_{\ell=1}\bigcup^\infty_{j\geq \ell}\Big\{|f_j|>\frac{1}{k}\Big\}\Big)=\lim_{\ell\rightarrow\infty}m\Big(\bigcup^\infty_{j\geq \ell}\Big\{|f_j|>\frac{1}{k}\Big\}\Big)\geq \lim_{\ell\rightarrow\infty}m\big(\Big\{|f_\ell|>\frac{1}{k}\Big\}\Big)$$
To complete the argument, notice that given $\varepsilon>0$, one can always find $k\in\mathbb{N}$ such that $\varepsilon<\frac1k$.

Continuity of $f$ implies that $\{f_j\leq\frac{1}{k}\}$ is a compact subset if ${0,1]$. on may try to leverage this along with compactness to get an almost measure-free proof of the result as proposed ny the OP.
