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
>  $$m\left ( \bigcup\limits\_{l=1}^{N} B\_l \right) \le 3^d\sum\limits\_{j=1}^{k} m(B\_{i\_j})$$
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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 $m(E\_0^c) < +\infty$.



If $\int_E f = 0$ for all $E$ the translation and dilation of $E_0$ then $f = 0 \text{ } a.e.$
The finite measure case is quite simple. Set $E\_\lambda = -\lambda E\_0$ for $\lambda > 0$, and $g\_\lambda = \mathbb{1}\_{E\_\lambda}$. By the assumption, $f \* g\_\lambda = 0$, which implies that $\hat{f} \hat{g}\_\lambda = 0$, where $\hat f$ is the Fourier transform of $f$. However, $\hat{g}\_\lambda(\xi) = \lambda^n \hat{g}\_1(\lambda \xi)$, and $\hat{g}\_1$ is continuous and positive at the origin, and hence non-zero in some ball $B(0, r)$. It follows that $\hat{f} = 0$ in $B(0, \lambda r)$ for every $\lambda > 0$, and consequently $f = 0$ almost everywhere.




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In the general case, we will need the following variant of Wiener's Tauberian theorem. By $\hat f$ we denote the Fourier transform of an integrable function $f$. The *spectrum* of a bounded function $g$ is the smallest closed set $S$ such that $g \* \varphi = 0$ for every Schwartz class function $\varphi$ such that $\hat\varphi(\xi) = 0$ for $\xi \in S$.



> 
> **Theorem:** If $g$ is a bounded function, $f$ is an integrable function, and $f \* g = 0$, then $\hat f = 0$ on the spectrum of $g$.
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For a proof, see Chapter 46 in:


* W.F. Donoghue Jr., *Distributions and Fourier Transforms*. Pure and Applied Mathematics, Vol. 32. Academic Press, New York–London, 1969.


Suppose that $f$ is an integrable function. The class $\cal F$ of sets $E$ such that $\int\_E f(x) dx = 0$ is a $\sigma$-algebra of sets.


If $\cal F$ contains every translation of a set $E\_0$, then for $E = -E\_0$ we have $f \* \mathbb{1}\_E = 0$. Hence, by Wiener's Tauberian theorem, $\hat f = 0$ on the spectrum $S$ of $\mathbb{1}\_E$. 


Suppose that $\cal F$ contains every translation of $\lambda E\_0 = -\lambda E$ for a given $E$ and every $\lambda > 0$. Since the spectrum of $\mathbb{1}\_{\lambda E}$ is $\lambda S$, we have $\hat f = 0$ on the union of $\lambda S$, where $\lambda > 0$.


We therefore have the following result.



> 
> **Proposition:** Suppose that $E\_0$ is a given set, and let $S$ be the spectrum of $\mathbb{1}\_{E\_0}$. Suppose that $f$ is an integrable function, such that $\int\_E f(x) dx = 0$ for every $E$ which is a translation of the dilation of $E\_0$. If $\bigcup\_{\lambda > 0} \lambda S$ is dense in $\mathbb{R}^n$, then necessarily $f = 0$. Conversely, if $\bigcup\_{\lambda > 0} \lambda S$ is not dense in $\mathbb{R}^n$, then there is a non-zero function $f$ with the above properties.
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The condition on the spectrum of $\mathbb{1}\_E$ is non-trivial: if $E\_0$ is the "checker-board set:
$$ E\_0 = \bigcup\_{i,j \in \mathbb{Z}} \bigl([2i, 2i+1) \times [2j, 2j+1)\bigr) \cup \bigl([2i+1, 2i+2) \times [2j+1, 2j+2)\bigr) , $$
then the spectrum of $\mathbb{1}\_{E\_0}$ is contained in $$(2 \pi \mathbb{Z} \times \{0\}) \cup (\{0\} \times 2 \pi \mathbb{Z}),$$
and hence $f$ need not be zero.


