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 $m(E_0^c) < +\infty$.


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.


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