Assume that we can find $\delta>0$, a subsequence $\{f_{n_k}\}$ and a sequence $\{x_{n_k}\}$ such that $f_{n_k}(x_{n_k})\geqslant \delta$ for all $k$. Bolzano-Weierstrass theorem ($[a,b]$ is compact) allows us to extract of $\{x_{n_k}\}$ a subsequence, denoted $\{t_j\}$, converging to some $t$. We have for all integers $n$ and $m$, denoting $g_k$ the sequence indexed by the integers appearing in $t_k$,
$$\delta\leqslant g_{m+n}(t_{m+n})\leqslant g_n(t_{m+n}).$$
Now we fix $n$, and take the $\limsup_{m\to +\infty}$. This gives for all integer $n$,
$$\delta\leqslant\limsup_{k\to +\infty}g_k(t_k)\leqslant g_n(t),$$
a contradiction.
In the second theorem, take $a=0, b=1$.
- If we don't assume $f$ continuous, consider $f_n(x):=x^n$.
- The same counter-example considering $(0,1)$.
Let $f:=\lim f_n$, and let $\varepsilon >0$ be given. We have to show that if $n$ is large enough, then $$\vert f_n(t)-f(t)\vert\leq\varepsilon\quad \hbox{for all $t\in[a,b]$}\, .$$
Since $f$ is assumed to be continuous, it is uniformly continuous on the compact interval $[a,b]$. So we may find a subdivision $a=t_0<t_1\dots <t_K=b$ of $[a,b]$ such that the oscillation of $f$ on each interval $[t_i,t_{i+1}]$ is less than $\varepsilon/2$.
Since $f_n(t_i)\to f(t_i)$ as $n\to\infty$ for $i=0,\dots ,K$, one can find $N$ such that if $n\geq N$, then $$\vert f_n(t_i)-f(t_i)\vert\leq\varepsilon/2\quad\hbox{for $i=0,\dots K$}\, .$$
Let us check that $\vert f_n(t)-f(t)\vert\leq \varepsilon$ for every $n\geq N$ and all $t\in [a,b]$.
Fix $n\geq N$, and take any $t\in [a,b]$. One can choose $i$ such that $t\in [t_i,t_{i+1}]$. Since the functions $f_n$ are non-decreasing we have $$f_n(t_i)\leq f_n(t)\leq f_n(t_{i+1})\, . $$
Since $\vert f_n(t_i)-f(t_i)\vert$ and $\vert f_n(t_{i+1})-f(t_{i+1})\vert$ are not greater than $\varepsilon/2$, it follows that $$f(t_i)-\varepsilon/2\leq f_n(t)\leq f(t_{i+1})+\varepsilon/2\, . $$
Moreover, since the oscillation of $f$ on $[t_i,t_{i+1}]$ is less than $\varepsilon/2$ and since $t\in [t_i,t_{i+1}]$, we also have $f(t_i)\geq f(t)-\varepsilon/2$ and $f(t_{i+1})\leq f(t)+\varepsilon/2$. Altogether, this gives $$f(t)-2\varepsilon/2\leq f_n(t)\leq f(t)+2\varepsilon/2\, , $$
which concludes the proof.
Best Answer
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 $m(U_n)< \epsilon$