$\{f_n\}$ be a sequence of continuous functions converging pointwise to $0$, show $\lim \int_0^1 f_n dx = 0$

For $\epsilon> 0$ and $n\ge 1$ define $$A_{n,\epsilon} \colon = \bigcap_{k\ge n} f_k^{-1}([0, \epsilon])$$ then $$A_{1,\epsilon} \subset A_{2,\epsilon} \subset \ldots$$ and $$\bigcup_{n} A_{n, \epsilon} = [0,1]$$

Consider a set $A_{n,\epsilon}$. For every $k \ge n$, the function $f_k$ takes values $\le \epsilon$ on $A_{n, \epsilon}$. By continuity, and the compactness of $A_{n,\epsilon}$, there exists a finite union of open intervals $U= U_{k,n, \epsilon}$ containing $A_{n, \epsilon}$ such that $f_k$ takes values $< 2 \epsilon$ on $U$. The complement of $U$ in $[0,1]$ is a finite union $E$ of closed intervals. We have $$\int_{[0,1]} f_k = \int_{\bar U}f_k + \int_{E}f_k \le 2 \epsilon + m(E)$$

Note that $E$ is an elementary subset (finite union of closed intervals) of $A_{n, \epsilon}^c$, an open subset of $[0,1]$.

Basic fact, proved below: If $U_n$ is a decreasing sequence of open subsets of $[0,1]$ with void intersection, and $E_n$ are elementary subsets of $U_n$ then $m(E_n) \to 0$. Once we prove this, we have the result.

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Define for an open subset $U$ of $[0,1]$, $m(U)\colon= \sup m(E)$, where $E$ is an elementary subset of $U$ ( so $m(U)$ is the interior Jordan measure of $U$).

One sees easily that $m$ is monotone: $m(U)\le m(V)$ if $U \subset V$.

For every $U$, and $\epsilon>0$, there exists $E_{\epsilon}\subset U$ elementary sucht that $m(E_{\epsilon}) > m(U) - \epsilon$. It's clear then that $m(U\backslash E_{\epsilon}) < \epsilon$.

Any elementary subset $E$ of $U\cup V$ is the union of elementary subsets $F$, $G$ of $U$, $V$, with intersections only at the boundary (Lebesgue covering lemma). Therefore, $m(U\cup V) \le m(U) + m(V)$.

Basic statement: If $U_n$ is a decreasing sequence of open subsets of $[0,1]$ with void intersection then $m(U_n) \to 0$.

Indeed, let $\epsilon > 0$. For every $n$ consider $E_n$ elementary such that $m(U_n \backslash E_n) < \epsilon/2^{n+1}$. Let $E'_n = E_1 \cap \ldots \cap E_n$. We have $$m(U_n \backslash E'_n) \le \sum_{k=1}^n m(U_n \backslash E_k) \le \sum_{k=1}^n m(U_k \backslash E_k)< \epsilon$$

Now $E'_n$ are elementary and form a decreasing sequence with empty intersection. By compactness, there exists $n$ such that $E'_n=\emptyset$. For that $n$ we have $m(U_n)< \epsilon$