- Suppose A is a Lebesgue measurable set in R with \(m ( A ) > 0\) Does there exists a sequence \(\{ x _ { n } \} _ { n = 1 } ^ { \infty }\) such that the complement of \(\textstyle { \bigsqcup _ { n = 1 } ^ { \infty } ( A + x _ { n } ) }\) in R has measure 0? Justify your answer. (We define \(A + x _ { n } = \{ a + x _ { n } \in \mathbb { R } : a \in A \} . )\)
Solution
1. Reduce to a bounded positive-measure subset.
Since \[ A=\bigcup_{R=1}^\infty \bigl(A\cap[-R,R]\bigr) \] and \(m(A)>0\), there is some \(R>0\) for which \[ B:=A\cap[-R,R] \] has positive finite measure. If countably many translates of \(B\) cover \(\mathbb R\) up to a null set, then the same translates of \(A\) do also. Thus it suffices to work with \(B\).
2. On a bounded interval, some translate captures a fixed proportion of any remainder.
Fix \(N\ge1\) and let \[ I_N=[-N,N]. \] Suppose \(E\subset I_N\) is measurable with \(m(E)>0\). By Tonelli’s theorem, \[ \begin{aligned} \int_{\mathbb R}m(E\cap(B+t))\,dt &=\int_E\int_{\mathbb R}\mathbf1_{B+t}(x)\,dt\,dx\\ &=m(E)m(B). \end{aligned} \]
If \(E\cap(B+t)\ne\varnothing\), then for some \(x\in[-N,N]\) and \(b\in[-R,R]\) we have \(x=b+t\), hence \[ t\in[-N-R,N+R]. \] Therefore the preceding integral is supported on an interval of length \[ L_N:=2(N+R). \] Consequently there exists \(t\in\mathbb R\) such that \[ m(E\cap(B+t)) \ge \frac{m(E)m(B)}{L_N}. \] Set \[ c_N:=\frac{m(B)}{L_N}>0. \] Then some translate of \(B\) captures at least the fraction \(c_N\) of \(E\).
3. Iteratively cover \(I_N\) up to a null set.
Set \(E_0=I_N\). If \(m(E_k)>0\), choose \(t_k\) as in Step 2 and define \[ E_{k+1}:=E_k\setminus(B+t_k). \] Then \[ m(E_{k+1}) \le (1-c_N)m(E_k). \] Inductively, \[ m(E_k)\le(1-c_N)^k m(I_N)\longrightarrow0. \] Since the remainders decrease, \[ m\left(I_N\setminus\bigcup_{k=0}^\infty(B+t_k)\right) =m\left(\bigcap_{k=0}^\infty E_k\right) =0. \] Thus countably many translates of \(B\) cover \(I_N\) modulo a null set.
4. Cover the whole line.
Perform Step 3 separately for each \(N=1,2,\dots\), and enumerate all the resulting translation parameters as one sequence \((x_n)\). Then for each \(N\), \[ m\left(I_N\setminus\bigcup_{n=1}^\infty(B+x_n)\right)=0. \] Because \[ \mathbb R=\bigcup_{N=1}^\infty I_N, \] the complement of \[ \bigcup_{n=1}^\infty(B+x_n) \] in \(\mathbb R\) is a countable union of null sets and is therefore null. Since \(B\subseteq A\), \[ \boxed{ m\left(\mathbb R\setminus\bigcup_{n=1}^\infty(A+x_n)\right)=0.} \] Hence the answer is yes.