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Since for a continuous random variable lets say, $X\sim N(0,1)$, we know that $P(X=x)=0$ for any $x\in\mathbb{R}$. I was wondering since we can extend probability measures over countable unions that $$P(X\in\mathbb{Q})=P\left(\bigcup_{x\in\mathbb{Q}}X=x\right)=\sum_{x\in\mathbb{Q}}P(X=x)=0$$ Thus we have that $X$ will almost never be rational thus implying that $P(X\in \mathbb{R}\backslash\mathbb{Q})=1$ meaning $X$ will almost surely be irrational. Is this analysis correct or am I doing something completely wrong?

Kamster
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You're right, this is a consequence of the fact that $\mathbb{Q}$ (or any other countable subset of $\mathbb{R}$) has measure zero. One can alternately show this by showing that all of $\mathbb{Q}$ can be covered by a union of intervals with arbitrarily small total length.

Travis Willse
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  • Ok, it just seems strange still hard for me to wrap head around it. Is there any intuition why things start to break down when trying to extend measure to arbitrary union of sets, rather than just Banach–Tarski paradox. – Kamster Oct 17 '14 at 07:53
  • I always wonder if we can't actually extend the measure to arbitrary union or you we only know for sure it works for countable – Kamster Oct 17 '14 at 08:09
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    The disjoint union/sum rule you're thinking of certainly doesn't hold for arbitrary unions. Otherwise, we could decompose any interval $[a, b]$, $a < b$, as $\bigcup_{x \in [a, b]} {x}$ and conclude that $[a, b]$ has measure zero. – Travis Willse Oct 17 '14 at 10:13