Sequence $f_n$ of continuous functions on $[0,1]$, such that $0 \le f_n \le 1$ and $\lim\limits_{n\to\infty}\int_{0}^{1} f_n(x)\,\mathrm dx=0$ Let $f_n$ be a sequence of continuous functions on $[0,1]$, such that $0 \le f_n \le 1$ and $$\lim_{n\to\infty}\int_{0}^{1} f_n(x)\,\mathrm dx=0.$$
a) The sequence ${f_n (x)}$ converges uniformly.
b) The sequence ${f_n (x)}$ converges for some $x$.
c) The sequence ${f_n (x)}$ might not converge for any $x$.
d) The sequence ${f_n (x)}$ converges for every $x$ but not necessarily uniformly.
The answer is c? I have no counter example in my hand.
 A: Consider the following sequence of sets, constructed as follows. Let $I_1=[0,1]$. Split $[0,1]$ in half, and let $I_2$ and $I_3$ be the first and second halves. Split $[0,1]$ in fourths, and let $I_4,I_5,I_6,I_7$ be the fourths in order. Continue this way, split $[0,1]$ in $2^n$ths and let $I_{2^n},\cdots, I_{2^{n+1}-1}$ be the $2^n$ pieces in order.
Now let $$f_n(x)=\text{ an isosceles triangle over } I_n \text{ of height } 1$$
Then $$\int_0^1 f_n\to 0$$ but if $x\neq m2^{-k}$ then $$\tag 1 \limsup_{n\to\infty} f_n(x)=1 \; ; \;\liminf_{n\to\infty} f_n(x)=0$$
ADD To fix the above, we can do the following: let $\langle g_n\rangle $ be the same as $\langle f_n\rangle $, but with the triangles shifted so that the peaks of the triangles fall on the binary rational points. Thus, $g_1$ will look like (the right) half an isosceles triangle on $[0,1/2]$ plus (the left) half of an isosceles triangle on $[3/4,1]$, $g_2$ will be an isosceles triangle over $[1/2,3/4]$, $g_3$ will be the two halves at $[0,1/4]$,$[7/8,1]$, and so on. Then $\langle g_1,f_1,g_2,f_2,g_3,f_3,\ldots\rangle$ should do: now the rational binary points oscillate between $0$ and $1$.
A: How about this:
$f_1=1$ everywhere;
$f_2=1$ on the left half of $[0,1]$ and $0$ on the right half; $f_3=1$ on the right half and $0$ on the left;
$f_4$, $f_5$, $f_6$ are respectively equal to $1$ on the left third, the middle third, and the right third and $0$ elsewhere;
$f_7$, $f_8$, $f_9$, $f_{10}$ are equal to $1$ respectively on the first, second, third, and fourth quarters, and $0$ elsewhere;
and so on.
Then $f_n(x)$ diverges for every  $x$ as $n\to\infty$, but $\displaystyle\int_0^1 f_n\to0$.
