TITLE: Is the sum of two irrational numbers almost always irrational?
QUESTION [9 upvotes]: Clearly the sum of two irrational numbers is not necessarily irrational. But is it true that it is 'almost always' irrational, in the sense that $$\displaystyle\lim_{x\to\infty}\dfrac{\lambda(P\cap B(x))}{\lambda(R\cap B(x))}=1$$ where $\lambda$ is Lebesgue measure, $R\subset \mathbb{R}^2$ is the set of points with irrational coordinates and $P\subset R$ is the set of those points the sum of whose coordinates is irrational (and $B(x)$ the disc of radius $x$ centred at the origin)? And I guess the same question applies to transcendental numbers. Intuitively it seems true, but I don't know how one would prove this. If not, then there is the question of whether the limit exists and if it does what is it.

REPLY [12 votes]: Let $NP$ be the set of pairs whose sum is rational. I think its easier to prove $\lambda(NP\cap B(x))=0$.  In fact since $NP\cap B(x)\subset NP$, we just prove $\lambda(NP)=0$ and we are done.  Let $$NP_x=\{(x,y)|x+y\in \mathbb{Q}\}$$
Notice that the restriction addition to this set is translation by $x$ which is measure preserving, hence the inverse image of the rational numbers has measure zero.  However, there is a weak form of Fubini's theorem that says that if a subset of a product measure space ( which $\mathbb{R}^2$ is) has the property that its intersection with each slice has measure zero then the set has measure zero. Hence $\lambda(NP)=0$.
To bring this back to the specific question you are asking, $NP\cup P=\mathbb{R}^2$, so for any open ball $B(x)$ $\lambda(P\cap B(x))=1$. 
On the other hand the set of points whose coordinates are irrational is the complement of a set of measure zero, so $\lambda(R\cap B(x))=1$.  Hence you are taking the limit of $1/1$.