$\{f_n\}$ be a sequence of continuous functions converging pointwise to $0$, show $\lim \int_0^1 f_n dx = 0$ This problem comes from Rudin's Real and Complex Analysis and is Exercise 2.10. Let $\{f_n\}$ be a sequence of continuous functions such that $0\leq f_n\leq 1$ and $f_n(x)\to0$ for all $x\in [0,1]$. We need to show that $$\lim_{n\to\infty}\int_0^1 f_n(x)dx = 0$$
In particular I am trying to show this without using Measure Theory or Lebesgue Integration.
What I have tried so far is to consider the sequence of functions $\{g_n\}$ defined by $$g_n(x)= \min\{f_1(x),\ldots f_n(x)\}$$
Then $\{g_n\}$ is a continuous, decreasing sequence of functions which converge pointwise to $0$. Since the sequence of continuous functions decreases to a continuous function, then we may apply Dini's Theorem to know that $g_n \to 0$ uniformly. Hence $$ \lim_{n\to 0}\int_0^1 g_n(x)dx = \int_0^1\lim_{n\to0}g_n(x)dx = 0$$
The only thing left to show is that $$ \lim_{n\to 0}\int_0^1 f_n(x)dx= \lim_{n\to 0}\int_0^1 g_n(x)dx $$
But this is where I am stuck. Again I am avoiding measure theory proofs.
 A: 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 $m(U_n)< \epsilon$
A: Isn't it a direct consequence of Lebesgue's Dominated convergence theorem ?
$(1)$ $f_n(x) \longrightarrow f(x) \equiv 0$ pointwise $\forall~x\in[0,1]$
$(2)$ For every $n\in\mathbb{N}$ , $|f_n(x)| < g(x) \equiv 1~~\forall~x \in[0,1]$
Therefore, by Lebesgue's DCT, $$\lim_{n\to\infty}\int_0^1 f_n(x)dx = \lim_{n\to\infty}\int_0^1 0~dx = 0$$
