# If $\int_E f = 0$ for all $E$ the translation and dilation of $E_0$ then $f = 0 \text{ } a.e.$

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 $$0 \frac{1}{n} ) < +\infty$$. 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 $$m(E_0^c) < +\infty$$.

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.

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 $$g$$.

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.

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.