Prove that following set is measurable and find its Lebesgue measure: $$ \{(x_1,\dots.x_n)\in\mathbb{R}^n:\sum_{i=1}^{n}|x_i|\notin\mathbb{Q};\forall_i|x_i|\leq1\} $$ I have only basic knowledge of Lebesgue measure (nothing about integrals). And I must admit, I do not feel I understand this basics well enough, so I would be thankful for any advice, which I am able to understand.
Find Lebesgue measure of $\{(x_1,\dots.x_n)\in\mathbb{R}^n:\sum_{i=1}^{n}|x_i|\notin\mathbb{Q};\forall_i|x_i|\leq1\}$
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measure-theory
lebesgue-measure
2 Answers
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Let $\lambda$ be the Lebesgue measure in $\Bbb R^n$. If $A = \{ (x_1,\dots,x_n) \in [-1,1]^n \mid \sum_{i=1}^n |x_i| \color{red} \notin \Bbb Q \}$ and $B = \{ (x_1,\dots,x_n) \in [-1,1]^n \mid \sum_{i=1}^n |x_i| \color{red} \in \Bbb Q \}$, then notice that $A \cap B = \emptyset$ and $A\cup B = [-1,1]^n$, which means that $\lambda(A) + \lambda(B) = \lambda([-1,1]^n) = 2^n$.
Notice that if $B_q = \{ (x_1,\dots,x_n) \in [-1,1]^n \mid \sum_{i=1}^n |x_i| = q \in \Bbb Q\}$ (possibly empty), then $B = \bigcup _{q \in \Bbb Q} B_q$ and $B_q \cap B_{q'} = \emptyset$ whenever $q \ne q'$, so $\lambda(B) = \sum _{q \in \Bbb Q} \lambda(B_q)$.
But $B_q$ is given by a real contraint imposed upon $n$ real variables, therefore it is an $n-1$-dimensional set in $\Bbb R^n$, and it is known that everything that has dimension $ To see that $A$ is measurable, it is enough to show that $B$ is measurable, because $A$ will then be the set difference of the measurable set $[-1,1]^n$ and the measurable set $B$, and set differences of measurable sets are measurable. To see that $B$ is measurable, it is enough to show that each $B_q$ is so, because $B$ will then be a countable union of the measurable sets $B_q$, and countable unions of measurable sets are measurable. Finally, to see that $B_q$ is measurable notice that $\{q\}$ is measurable in $\Bbb R$, therefore consider the function $f : [-1,1]^n \to \Bbb R$ given by $f(x_1, \dots, x_n) = \sum _{i=1} ^n |x_i|$. This is a continuous function, therefore measurable, therefore $B_q = f^{-1} (\{q\})$ is measurable (because preimages of measurable sets under measurable maps are measurable).
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Note that for each $q \in \mathbf Q_{\ge 0}$, and every choice of signs $\epsilon_i \in \{\pm 1\}$, the set $$ H_{\epsilon, q} = \left\{x \in \mathbf R^n : \sum_i \epsilon_i x_i = q \right\} $$ is a hyperplane, and hence $\lambda(H_{\epsilon,q}) = 0$. As $$ \left\{x \in \mathbf R^n : \sum_i |x_i| \in \mathbf Q\right\} \subseteq \bigcup_{\epsilon \in \{\pm 1\}^n, q\in \mathbf Q} H_{\epsilon, q} $$ and a countable union of null-sets is a null-set, we have $$ \lambda\left(\left\{x \in \mathbf R^n : \sum_i |x_i| \in \mathbf Q\right\}\right) = 0 $$ Therefore $$ \lambda\left(\left\{x \in \mathbf R^n : \sum_i |x_i| \not\in \mathbf Q, \|x\|_\infty \le 1\right\}\right) = \lambda([-1,1]^n) = 2^n $$