Let $f \in L^1(\mathbb{R}^n)$. It's obvious that if $\int_R f = 0$ for all rectangles $R$ then $f = 0$ $a.e.$ since every open set is union of almost disjoint rectangles and consequently with zero integration. Then I came up with a question. Can rectangles be replace by balls? The answer is yes by elenebtary version of a **Vitali covering argument**

Suppose $\mathcal{B} = \{ B_1, \dots , B_N\}$ is a finite collection of open balls in $\mathbb{R}^d$. Then there exists a disjoint sub-collection $B_{i_1}, \dots , B_{i_k}$ of $\mathcal{B}$ that satisfies $$m\left ( \bigcup\limits_{l=1}^{N} B_l \right) \le 3^d\sum\limits_{j=1}^{k} m(B_{i_j})$$

If $m(f \neq 0 ) >0$ then there is $n$ such that $0<m( f > \frac{1}{n} ) < +\infty$. So there is $F \subset E \subset G, m(G\setminus F)<\epsilon$ for arbitrary $\epsilon > 0$ and with $F$ compact and $G$ open. Therefore there is $\mathcal{B} = \{ B_1, \dots , B_N\}$ with$F \subset \cup \mathcal{B} \subset G$ and disjoint sub-collection $B_{i_1}, \dots , B_{i_k}$ of $\mathcal{B}$ that satisfies $$m\left ( \bigcup\limits_{l=1}^{N} B_l \right) \le 3^d\sum\limits_{j=1}^{k} m(B_{i_j})$$ Therefore $$\sum\limits_{j=1}^{k} \int_{B_{i_j}} f \ge \frac{3^{-d}}{2n}m(f> \frac{1}{n})$$ with small $\epsilon$, a contradiction.

Then I go with much general situation considering bounded measurable set $E_0$ with positive measure and its translations and dilations. This is easy to do. Let $E_0 \subset B_0$ with $B_0$ a ball. Then put similar $E_i$ in the ball $B_i$ with the same scale as $E_0$ and $B_0$ in the above situation so we have $$\frac{m(B_0)}{m(E_0)}\sum\limits_{j=1}^{k} \int_{E_{i_j}} f \ge \frac{3^{-d}}{2n}m(f> \frac{1}{n})$$

Now come to the ultimate situation that I can't work out. **What if $E_0$ is simply measurable with positive measure(or more simple case, finite positive measure)?**

Just assume $m(E_0^c) = +\infty$. If $m(E_0^c) < +\infty$ then contract $E_0$ by scale $\delta$ to make $m(\delta E_0^c) \approx 0$. Then we have $\int_{\mathbb{R}^n} f = 0$. Therefore $\int_{E_0^c} f = \int f - \int_{E_0} f = 0$ with $m(E_0^c) < +\infty$.