Let $n≥2$ and $$G:=\{a+tx\mid t\in\mathbb{R}\}, a\in\mathbb{R}^n, x\in\mathbb{R}^n-\{0\}$$ $$F:=\{G\subset\mathbb{R}^n\mid\exists x,y,\in\mathbb{Q}^n\cup G:x\neq y\}$$Let $F$ be the set that contains all straight lines, $G$, with two rationals. Show that $$M:=\cup_{G\in F}G\subset\mathbb{R}^n$$ is Lebesgue-measurable and determine the measure $\mu(M)$.
In the assignment before this I have shown that for every straight line in $\mathbb{R}^n, n≥2$ that $\mu(G)=0$. How can I use this result here?
I'd appreciate any help.
in your definition of $F$ you mean $\cap$, not $\cup$. The set of all rational numbers is countable. It follows $\Bbb Q^n$ is also countable, and it follows that the family $F$ of lines through two rationals (or rather elements of $\Bbb R^n$ with all coordinates rational) is countable. Lebesgue measure is countably additive, so the union of countably many lines, each of measure $0$ must itself be of measure $0$.
In case the term "countably additive" is not clear, here is a more detailed proof. The family $F$ could be written as a list of lines: $F=\{l_n:n\in\Bbb N\}$ (where $\Bbb N$ is the set of all positive integers). Take $\varepsilon>0$. For each $n$ take an open set $U_n$ containing the line $l_n$ such that $\mu(U_n)<\varepsilon/2^n$. Then $F\subset\cup_{n\in\Bbb N}\,U_n$, hence $\mu(F)\le\mu(\cup_{n\in\Bbb N}\,U_n)\le\sum_{n\in\Bbb N}\mu(U_n)\le\sum_{n\in\Bbb N}\varepsilon/2^n=\varepsilon$. Since $\mu(F)\le\varepsilon$ for all $\varepsilon$, it follows $\mu(F)=0$.