$S ⊂\mathbb{R}$ is 1 periodic if $x ∈S$ iff $x + 1 ∈S$. Show that
if $S$ is 1-periodic and there exists a countable set $β_1,…$ so that $S + β_i ∩S + β_j = ∅$ for all $i\neq j$ and $\bigcup\limits_{j=1}^{\infty}(S + β_j) = \mathbb{R}$,
then $S$ is not Lebesgue measurable.
I want to assume $S$ is Lebesgue measurable and derive a contradiction.
consider $S \cap[0,1]$ and $N_{\beta_j}=\{x+\beta_j \mid x \in S \cap [0,1-\beta_j]\} \cup \{x+\beta_j \mid x \in S \cap [1-\beta_j,1]\}$.
Then $[0,1]=\bigcup\limits_{j=1}^{\infty}N_{\beta_j}$ so that $\mu([0,1])=\sum\limits_{j=1}^{\infty}\mu(N_{\beta_j})=\sum\limits_{j=1}^{\infty}\mu(S \cap [0,1])=\infty$ a contradiction.
Best Answer
Let us assume that $S$ is measurable.
Consider the sets $$T_n := (S + \beta_n) \cap [0, 1).$$ Also define $T := S \cap [0, 1)$. Then, each $T_n$ and $T$ are measurable as well.
Note that we have some quite direct consequences of our assumptions on $S$:
There is one tricky bit: $\mu(T_n) = \mu(T)$ for all $n \geqslant 1$, i.e., all of them have the same measure.
From this, the contradiction is immediate, since we $$1 = \mu([0, 1)) = \sum_{n = 1}^{\infty} \mu(T_n) = \infty \cdot \mu(T).$$
Thus, it comes down to showing the tricky bit (we haven't used periodicity of $S$ yet).
Fix $n \geqslant 1$. Note that since $S$ is invariant under translations by integers, we may assume that $\beta_n \in [0, 1)$.
We may further assume $\beta_n \neq 0$ else $T_n = T$ anyway.We don't need this assumption. It's just that some intervals below may be empty but that is okay.For ease of notation, let us write $A \approx B$ to mean $\mu(A) = \mu(B)$.
We have \begin{align} T_n &= (S + \beta_n) \cap [0, 1) \\ &= \left((S + \beta_n) \cap [0, \beta_n)\right) \sqcup \left((S + \beta_n) \cap [\beta_n, 1)\right) \\ &\approx \left((S + \beta_n) \cap [1, 1 + \beta_n)\right) \sqcup \left(S \cap [0, 1 - \beta_n)\right). \end{align} To see the last step, note that periodicity of $S$ tells us that the intersection of $S + \beta_n$ has the same measure when intersected with $[a, b)$ or $[1 + a, 1 + b)$. This explains why the first sets in the last two lines have the same measure. The last sets are clearly just translates of each other. Now, we can further translate the first set to get \begin{align} T_n &\approx \left((S + \beta_n) \cap [1, 1 + \beta_n)\right) \sqcup \left(S \cap [0, 1 - \beta_n)\right) \\ &\approx \left(S \cap [1 - \beta_n, 1)\right) \sqcup \left(S \cap [0, 1 - \beta_n)\right) \\ &= T, \end{align} as desired.