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Let $A \subseteq \Bbb{R}$ be Borel measurable, and $T$ a dense subset of $\Bbb{R}$. Suppose for every $t \in T$ that $$m((A+t)\setminus A)=0,$$

where $m$ is the Lebesgue measure. Then I want to show that $m(A)$ or $m(\Bbb{R}\setminus A) = 0$.

Indeed suppose $m(A)$ is finite. Fix $t \in T$, then we see that $$\chi_A(x)= \tau_t \chi_A(x)$$ for a.e. $x$, where $\tau_t$ is translation. Then taking Fourier transforms we get $$\hat{\chi_A}(\xi) = e^{it\xi} \hat{\chi_A}(\xi).$$ Since $t \in T$ was arbitrary, given any $\xi$ we can always choose $t\in T$ so that $e^{it\xi} \neq 1,$ and we get the result that we want.

My question is: How can we reduce to the case where $m(A) < \infty$?

Silvia Ghinassi
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David Benjamin Lim
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  • Do you mean $(A+t)\setminus A $ or rather $(A+t)\Delta A $ (symmetric difference)? – PhoemueX Feb 10 '16 at 08:03
  • The result does not hold for the counting measure on Z^+. Perhaps a direct elementary proof for Lebesgue measure on R would show where the finiteness of m(A) affects the argument . – DanielWainfleet Feb 10 '16 at 13:11
  • It seems to me that with some little adaptation, the proof given here does work in this case too. – Tomás Feb 10 '16 at 20:12
  • Very nice your idea to use Fourier transform to prove the case where $m(A)$ is finite. In fact, under the conditions of the result, if $m(A) < \infty$ then $m(A) =0$. However, I don't think there is an easy or natural way to reduce the general case to the case where $m(A) < \infty$. I offer you a different (complete) proof of the result (see my answer below). – Ramiro Feb 11 '16 at 16:15

1 Answers1

1

Let $A \subseteq \Bbb{R}$ be Borel measurable, and $T$ a dense subset of $\Bbb{R}$. Suppose for every $t \in T$ that $$m((A+t)\setminus A)=0,$$ where $m$ is the Lebesgue measure. Then $m(A)=0$ or $m(\Bbb{R}\setminus A) = 0$.

Proof:

We define $-A$ as $-A=\{-x : x \in A\}$.

Note that for all $t, x \in \Bbb{R}$

$$\chi_{A+t}(x) = \chi_{A}(x-t)=\chi_{-A+x}(t) $$

Note also that, for all $t, x \in \Bbb{R}$,

$$\chi_{(A+t)\setminus A}(x)= \chi_A(x-t) - \chi_A(x-t)\chi_A(x)$$

Note also that, by Lebesgue measure properties, for all $t, x \in \Bbb{R}$, $$m(A)=m(A+t)=m(-A)=m(-A+x)$$

I will divide the rest of the proof in two parts

Part 1. Let $t\in \Bbb{R}$ and, since $T$ a dense subset of $\Bbb{R}$, let $\{t_n\}_n$ be a sequence such that, for all $n$, $t_n \in T$ and $\{t_n\}_n$ converges to $t$. Then we have that $\chi_A(x-t_n) - \chi_A(x-t_n)\chi_A(x)$ converges pointwise to $\chi_A(x-t) - \chi_A(x-t)\chi_A(x)$.

Since the functions $\chi_A(x-t_n) - \chi_A(x-t_n)\chi_A(x)$ are non-negative, we can apply Fatou's lemma and we get: \begin{align} 0 \leqslant m((A+t)\setminus A) &= \int \chi_{(A+t)\setminus A}(x) dx = \int (\chi_A(x-t) - \chi_A(x-t)\chi_A(x)) dx = \\ &= \int \liminf_n (\chi_A(x-t_n) - \chi_A(x-t-n)\chi_A(x)) dx \leqslant \\ &\leqslant \liminf_n \int (\chi_A(x-t_n) - \chi_A(x-t-n)\chi_A(x)) dx = \\ &= \liminf_n \: m((A+t_n)\setminus A)=0 \end{align} So, we proved that, for all $t \in \Bbb{R} $, $m((A+t)\setminus A)=0$.

Part 2: Since, for all $t \in \Bbb{R} $, $m((A+t)\setminus A)=0$, we have, applying Tonnelli's theorem:
\begin{align} 0 &= \int m((A+t)\setminus A) \;dt = \int \int (\chi_A(x-t) - \chi_A(x-t)\chi_A(x)) \;dx \; dt = \\ &= \int \int (\chi_A(x-t) - \chi_A(x-t)\chi_A(x)) \;dt \; dx = \\ &= \int \int (\chi_{-A+x}(t) - \chi_{-A+x}(t)\chi_A(x)) \;dt \; dx = \\ &= \int \int \chi_{-A+x}(t)(1 - \chi_A(x)) \;dt \; dx = \\ &= \int m(-A+x)(1 - \chi_A(x)) \;dx = \\ &= \int m(A)(1 - \chi_A(x)) \;dx = \\ &= m(A)\int ( 1 - \chi_A(x) ) \; dx = m(A)m(\Bbb{R}\setminus A) \end{align} So we have that $m(A)=0$ or $m(\Bbb{R}\setminus A) = 0$.

Ramiro
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