Sequence of functions without pointwise convergence Is there a sequence of continuous non negative functions on $[0,1]$, such that  $ \lim_{n\to \infty} \int_{0}^{1} f_n(x)$ is $0$, and $f_n(x) $ doesn't have a pointwise convergence?  The best I could do I come up with functions satisfying integral condition but the no pointwise convergence condition is missing. 
 A: Let's write, for integers $n\geq1$ and $1\leq k\leq n$: $$\chi_{n,k}(x)=\begin{cases}1&\text{if }(k-1)/n<x\leq k/n\\ 0&\text{otherwise}\end{cases}.$$ Now, set $f_{(n-1)n/2+k}=\chi_{n,k}$ for each $n\geq 1$ and $1\leq k\leq n$ (you can check that this actually counts from $1, 2,\ldots$). For $n\geq 1$, $1\leq k\leq n$, we have that $\int_{0}^{1}f_{(n-1)n/2+k}\mathrm{d}x=1/n$, so we certainly have that the limit of the integrals is 0, but at any point $x\in(0,1]$, the sequence $(f_{j}(x))_{j=1}^{\infty}$ takes the values 0 and 1 infinitely many times, so $f_{j}(x)$ cannot converge for any $x$ in this range.
To get continuity, let's "tie down the ends" of the $\chi_{n,k}$ (call these $\tilde{\chi}_{n,k}$) so that in the neighboring intervals to $[(k-1)/n,k/n],$ namely $[(k-2)/n,(k-1)/n]$ and $[k/n,(k+1)/n]$ (omitting one or the other of these if $k=1$ or $k=n$), the function is linear going from 0 to 1 if the left neighbor, and linear going from 1 to 0 if the right neighbor. This makes the $\tilde{\chi}_{n,k}$ (and thus the analogously-defined $\tilde{f}_{j}$) continuous, and $\int_{0}^{1}\tilde{f}_{(n-1)n/2+k}(x)\mathrm{d}x\leq 3/n$, so the integrals still tend to 0. Since we retain the property that for all $x\in(0,1]$, $(\tilde{f}_{j}(x))_{j=1}^{\infty}$ takes the values 0 and 1 infinitely many times, these sequences do not converge. Note also that we should set $\tilde{\chi}_{n,1}(0)=1$ in this case, so that we have continuity up to the left endpoint of the interval, and since $\tilde{\chi}_{n,k}(0)=0$ whenever $k>1$, $(\tilde{f}_{j}(0))_{j=1}^{\infty}$ also fails to converge.

