# Outer Measure of the set $E=(\mathbb Q\times \mathbb R) \cup (\mathbb R\times \mathbb Q)$

I need to find the Lebesgue outer measure value of the following set: $$E=(\mathbb Q\times \mathbb R) \cup (\mathbb R\times \mathbb Q)$$. However, I do not know where to start.

My attempt:

I know that $$\mathbb Q$$ is countable so we can enumarate it as follows $$\mathbb Q=\{x_1,x_2...\}$$. Hence, around each point $$x_i$$ we have an interval $$(x_i-\frac{\epsilon}{4^k},x_i+\frac{\epsilon}{4^k})$$. I want to show that E can be covered by open cells of the form $$(x_i-\frac{\epsilon}{4^k},x_i+\frac{\epsilon}{4^k})\times (x_j-\frac{\epsilon}{4^k},x_j+\frac{\epsilon}{4^k})$$ and somehow show that the measure is $$0$$ i.e., I do not know if that is the actual value.

Anyway any hints or solutions are welcome.

Using your enumeration $$\mathbb{Q} = \{x_1, x_2, \cdots\}$$, for each $$\epsilon > 0$$ consider the following sets:
$$\begin{gathered} A_{i,n} = (x_i - \epsilon 4^{-i-n}, x_i + \epsilon 4^{-i-n}) \times (-2^n, 2^n), \\ B_{i,n} = (-2^n, 2^n) \times (x_i - \epsilon 4^{-i-n}, x_i + \epsilon 4^{-i-n}). \end{gathered}$$
It is easy to see that $$E \subseteq \bigcup_{i,n\geq 1} A_{i,n} \cup B_{i,n}$$. Moreover, if $$m^*$$ denotes the outer Lebesgue measure, then
\begin{align*} m^*(E) &\leq \sum_{i,n \geq 1} (\operatorname{Area}(A_{i,n})+\operatorname{Area}(B_{i,n})) \\ &= \sum_{i,n \geq 1} 2 \cdot (2 \cdot \epsilon 4^{-i-n}) \cdot (2 \cdot 2^n) \\ &= \sum_{i,n \geq 1} \epsilon 2^{3-2i-n} \\ &= \frac{8}{3}\epsilon, \end{align*}
which can be made arbitrarily small. Therefore $$m^*(E) = 0$$.
You can write $$\mathbb{Q}\times \mathbb{R}$$ as countably many strips, i.e. $$(\mathbb{Q}\times \mathbb{R}) \cup (\mathbb{R}\times \mathbb{Q}) = (\cup_{n\in\mathbb{N}} \{q_n\}\times \mathbb{R}) \cup (\cup_{n\in\mathbb{N}} \mathbb{R}\times \{q_n\}),$$ where $$\{q_n\}_{n\in\mathbb{N}} = \mathbb{Q}$$. Use countable sub-additivity of the outer measure and the fact that $$\mathcal{L}^2(\{x\}\times \mathbb{R})=0$$ for all $$x\in \mathbb{R}$$.