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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.

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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 $m^*(E) = 0$.

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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 $x\in \mathbb{R}$.

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