Functions sequence and lebesgue measure I'm stuck on a problem which seems quite difficult.
I need a functions sequence $(f_n)$ (each fn is continuous) defined in $[0,1] ×[0,1] $ such that $(f_n)$ doesn't converge puntually in any point but
$$\int fn\ \rightarrow 0$$
If anyone can help me I would be thankful. Any idea can be helpful.
It should work if you first try to find a decent discontinuous example for a function defined in $[0,1]$.
What you can do is create a sequence of subsets of $[0,1]$ that exists of intervals of decreasing length that repeatedly cover up the whole interval.
For example intervals of the form $[\frac{k-1}{m},\frac{k}{m}]$ for $m=1,2,3...$ and $k=1,2,...,m$ should do the job if you numerate them like $I_1=[0,1]$, $I_2=[0,\frac{1}{2}]$, $I_3=[\frac{1}{2},1]$, $I_4=[0,\frac{1}{3}]$, $I_5=[\frac{1}{3},\frac{2}{3}]$ and so on.
If you now take the squence of characteristic functions $\chi_{I_n}$ the sequence is not pointwise convergent for any point on the Interval $[0,1]$ but satisfies $\lim\limits_{n\to\infty}\int_{[0,1]}\chi_{I_n}\mathrm{d}\mathcal{L}^1=0$.
You can then try to make the functions continuous by defining them on a ball slightly greater than the original interval for example $\tilde{I}_n:=[\min(I_n)-\frac{1}{n}, \max(I_n)+\frac{1}{n}]$.
Now you define
\begin{equation}
f_n(x):=\begin{cases} 1 &amp; \text{ if } x\in I_n\\
0 &amp; \text{ if } x\in\tilde{I}_n^c\\
nx-n\min(\tilde{I}_n) &amp; \text{ if } x\in[\min(\tilde{I}_n),\min(\tilde{I}_n)+\frac{1}{n}]\\
-nx+n\max(\tilde{I}_n) &amp; \text{ if } x\in[\max(\tilde{I}_n)-\frac{1}{n}, \max(\tilde{I}_n)]
 \end{cases}
\end{equation}
Now the only thing left to do is take $\tilde{f}_n:=f_n\big|_{[0,1]}$ and you are done.
Now to get a sequence of functions defined on $[0,1]^2$ you take $g_n(x,y):=\tilde{f}_n(x)\chi_{[0,1]}$.
This sequence should then satisfy all your wanted conditions.
I'd recommend drawing the functions if you have problems picturing them.
The idea is that you get trapezoid-like shapes that have a volumes that converge to $0$.
There is also probably a simpler way to do this but this is at least what I could come up with so let me know if it helped.
I also just assumed that you where looking for positive functions.
Choose $f_n(x, y) := \sqrt{n}\cos(nx)$. Then, using Fubini:
$$
\int_{[0, 1] \times [0, 1]} f_n(x, y)~\mathrm{d}(x, y) = \int^1_0 \int^1_0 \cos(nx)~\mathrm{d}y~\mathrm{d}x = \int^1_0  \cos(nx)~\mathrm{d}x = \frac{1}{\sqrt{n}}\sin(n) \rightarrow 0
$$
But note that for all $(x, y) \in [0, 1] \times [0, 1]$, the sequence $( \sqrt{n}\cos(nx))_{n \in \mathbb{N}}$ diverges
.