How can I show that one of $m(A)$ or $m(\Bbb{R}\setminus A)$ is zero? 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 $m(A) < \infty$?

 A: 
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 $m(\Bbb{R}\setminus A) = 0$.
