As part of a proof it is stated (without explanation) that the Lebesgue measure of a $x \in\mathbb{R}^n\setminus\mathbb{Q}^n$ is infinite.
I cannot see why? Can somebody explain?
Why is the lebesgue measure of an irrational vector infinte?
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measure-theory
lebesgue-measure
outer-measure
1 Answers
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There are different ways to explain this. Easiest is to notice that Leb $\mathbb{Q}^n=0$, for all $n=1,2,\ldots$. Since $\mathbb{Q}^n$ is countable, we can express each $p_i\in \mathbb{Q}^n$ as $$p_i=(q^i_1,q^i_2,\ldots ,q^i_n),$$ where $i=1,2,\ldots$ and $q_j^i\in \mathbb{Q}$ for $j=1,2,\ldots ,n$ and knowing that Lebesgue measure of each rational number is $0$: \begin{align*} \text{Leb }\mathbb{Q}^n&=\text{Leb } p_1+\text{Leb } p_2+\ldots\\ &=\text{Leb }q^1_1+\ldots +\text{Leb }q^1_n+\text{Leb }q_1^2+\ldots +\text{Leb }q_n^2+\ldots \\ &=0. \end{align*} Since Leb $\mathbb{R}^n=\infty$, it follows: Leb $\mathbb{R}^n\setminus \mathbb{Q}^n=\infty$.