$ \mu(\{x\in A :\, x-\epsilon\not\in A\})\rightarrow 0$?

Question

Let $A \subset \mathbb{R}$ be a measurable set. Consider $$ A_\epsilon=\{x\in \mathbb{R} :\, x-\epsilon\in A\}\, ,$$ I guess it should be true that, for $\epsilon \rightarrow 0$, and $$ \mu( A \setminus A_\epsilon)\rightarrow 0 $$ but I can't prove it... Any hint?

Answer 1

The assertion does not hold true. Consider

$$A := \bigcup_{k \in \mathbb{N}} [2k,2k+1],$$

then

$$A_{\epsilon} = \epsilon + A = \bigcup_{k \in \mathbb{N}} [2k+\epsilon,2k+1+\epsilon]$$

which implies

$$A \backslash A_{\epsilon} = \bigcup_{k \in \mathbb{N}} [2k,2k+\epsilon)$$

for $\epsilon \in (0,1)$. Consequently,

$$\mu(A \backslash A_{\epsilon}) = \sum_{k \in \mathbb{N}} \mu([2k,2k+\epsilon)) = \infty$$

for any $\epsilon \in (0,1)$.