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 $m(U_n)< \epsilon$
Rather than give formulas, I will describe the graphs of the functions, which are piecewise linear. Think of a “discretely moving wave”, where after the wave is done moving, we switch to a “thinner” wave.
Let $f_1$ go from $(0,1)$ to $(\frac{1}{2},0)$ to $(1,0)$. Let $f_2$ go from $(0,0)$ to $\frac{1}{2},1)$ to $(1,0)$. Let $f_3$ go from $(0,0)$ to $\frac{1}{2},0)$ to $(1,1)$.
Then let $f_4$ go from $(0,1)$ to $(\frac{1}{4},0)$ to $(1,0)$. Let $f_5$ go from $(0,0)$ to $(\frac{1}{4},1)$ to $(\frac{1}{2},0)$ to $(1,0)$. Then $f_6$ from $(0,0)$ to $(\frac{1}{4},0)$ to $(\frac{1}{2},1)$ to $(\frac{3}{4},0)$ to $(1,0)$. Then $f_7$ go from $(0,0)$ to $(\frac{1}{2},0)$ to $(\frac{3}{4},1)$ to $(1,0)$. Then $f_8$ from $(0,0)$ to $(\frac{3}{4},0)$ to $(1,1)$.
Next do the same thing with the partition $0\lt \frac{1}{8}\lt\frac{1}{4}\lt\frac{3}{8}\lt \frac{1}{2}\lt \frac{5}{8}\lt\frac{3}{4}\lt\frac{7}{8}\lt 1$. Then partitioning $[0,1]$ into $16$ equal subintervals, etc.
Each new batch of functions has a smaller integral, all positive, but converging to $0$. The functions are all continuous. And $f_n(x)$ does not converge for any $x$ because you can always find arbitrarily large values of $n$ where $f_n(x)=0$ and values where $f_n(x)$ is very close to $1$ (by approximating $x$ with a rational with denominator a power of $2$).
Best Answer
(second question) $f_n(x)=\frac1n\sin(n^2x)$
(third question) $f_n(x)=1/x$ on $[1/n,1]$ and $n^2x$ on $[0,1/n].$