A problem on sequence of functions on $[0,1]$ Does there exist a sequence of functions $f_n : [0,1] \to [0, \infty)$ such that $$\lim_{n \to \infty} \int\limits_{0}^{1} f_n(x) dx = 0$$ but there doesn't exist any $ x \in [0,1] $ for which the sequence $f_n(x)$ converges ?
Can anyone guide to how to solve these types of problems. Thank you.
 A: I dont know a way to "guide" you, you only need to think about it.
We need some sequence of integrable functions such that the limit $\lim_{n\to\infty}f_n(x)$ doesn't exists for all $x\in[0,1]$. There is a lot of ways to ensure that this limit doesn't exist, by example making $f_n(x)$ frequently zero and frequently one for each $x\in[0,1]$.
By the other hand you need that the limit of the sequence of integrals will be zero. One way to do it is, by example, that each integrand will be an indicator function of a set of decreasing measure.
Putting these two ideas together and assuming that we are working with the integral of Riemann we can set $f_n:=\mathbf{1}_{I_n}$ for a sequence of intervals $(I_n)$ that moves a fixed distance from left to right when $n$ increases, and that when it reach the right hand side of the interval $[0,1]$ then it moves again to the left endpoint of the interval $[0,1]$ but decreasing it length and the distance of translation by some amount, and so on.
This is a kind of typewriter sequence of intervals (each interval seems like typing a word in a typewriter, and when we reach the final of a phrase we move again to the first place). By example we can set the sequence of intervals $I_{k,j}$ defined by
$$
I_{k,j}:=\left[\frac{j}{k},\frac{j+1}{k}\right),\quad k\in \Bbb N_{> 0} ,\,j\in\{0,\ldots ,k-1\}
$$
After you can map the set of pairs $(k,j)$ bijectively to $\Bbb N $ such that it respect the lexicographic order of the pairs $(k,j)$ defined by
$$
(a,b)\leqslant(a',b ')\iff a '>a \,\lor\, (a '=a \,\land\, b'>b)
$$
This ensures that the intervals are in the order that represent what we want to do. In our example this bijection is the map $(k,j)\mapsto \frac{(k-1)k}2+j+1$ as you should verify.
