The Lebesgue point theorem for locally average function Let $Q=(0,1)\times(0,1)\subset \mathbb R^2$ and $u\in L^\infty(Q)$ be given. Define, for $N\in\mathbb N$, that a small cube
$$
Q_{N}(i,j):=(i/N,(i+1)/N)\times (j/N,(j+1)/N)
$$
where $0\leq i,j<N$. Hence, $Q=\bigcup_{0\leq i,j<N}Q_N(i,j)$ and each $Q_N$ are mutually disjoint.
My question: do we have 
$$
\lim_{N\to\infty}\sum_{0\leq i,j<N}\left\|u-\frac{1}{|Q_N(i,j)|}\int_{Q_{N}(i,j)}u(y)dy\right\|_{L^2(Q_N(i,j))}=0
$$
hold?
At the beginning I though this is just an easy conclusion from Lebesgue point theorem, but later I realize it may not be that easy...
Any help is really welcome!
 A: I think this proof using lusin's theorem works. I still think there is a proof using a variation of Lebesgue point theorem and dominated convergence, see my comment.
Let say we're in $[0;1]$ instead of $[0;1]^2$, it doesn't change the proof much and it simplify the notations.
First your result is true if $f$ is continuous, by the dominated convergence theorem for example. Now take a function $f$ in $L^\infty[0;1]$, let's call $f_n$ the sequence of local average approximation of $f$ you use in your question. Without lost of generality assume that $|f(x)|\leq M$ everywhere (and not only almost everywhere). We use Lusin's theorem : for every $\varepsilon>0$ there exists a compact $K$ and a continuous function $g$ such that $g_{|K}=f_{|K}$, $\|g\|_\infty\leq M$ and such that $\lambda(K)=1-\varepsilon$. We also call $g_n$ the sequence of local average approximations of $g$, one has that $g_n\to g$ in $L^2[0;1]$.
Now we use the so called "$3\varepsilon$ trick" :
$$\|f_n-f\|_2\leq \|f_n-g_n\|_2+\|g_n-g\|_2+\|g-f\|_2 $$
We have $||g-f||_2\leq 2\sqrt{\varepsilon} M$ and for $n$ large enough we also have $\|g_n-g\|_2\leq \varepsilon$. Now we evaluate the last term :
$$\|f_n-g_n\|_1=\int|f_n-g_n|d\lambda=\sum_{k=1}^n\int_{\frac{k-1}{n}}^{\frac{k}{n}}\left|n\int_{\frac{k-1}{n}}^{\frac{k}{n}}f-g d\lambda\right|d\lambda$$$$\leq\sum_{k=1}^n n\int_{\frac{k-1}{n}}^{\frac{k}{n}}\int_{\frac{k-1}{n}}^{\frac{k}{n}}|f-g| d\lambda d\lambda=\sum_{k=1}^n \int_{\frac{k-1}{n}}^{\frac{k}{n}}|f-g|d\lambda=\|f-g\|_1\leq 2\varepsilon M.$$
Since the $f_n-g_n$ are bounded by $2M$ uniformly in $n$ we have 
$$\|f_n-g_n\|_2^2\leq 2M \|f_n-g_n\|_1\leq 4M^2\varepsilon$$ and thus $\|f_n-g_n\|_2\leq 2M\sqrt\varepsilon$
Combining all this we get that, for $n$ large enough, $\|f-f_n\|_2\leq 4M\sqrt\varepsilon +\varepsilon$. Since $\varepsilon$ was arbitrary this conclude the proof.
Edit here is the proof in the case where $f$ is continuous. Since the $f_n$ are all uniformly bounded by $M$ to apply dominated convergence we only need to show pointwise convergence of the $f_n$. Take $x\in [0;1]$ and $\varepsilon>0$. There is a $\delta>0$ such that $|x-y|<\delta\Rightarrow |f(x)-f(y)|< \varepsilon$. For every $n$ such that $1/n<\delta$ the number $x$ is contained in a $I_n=[k_n/n;(k_n+1)/n[$ interval and thus
$$f_n(x)=n\int_{I_n} f(x) dx\in[f(x)-\varepsilon;f(x)+\varepsilon] .$$
In other words $|f_n(x)-f(x)|<\varepsilon.$
