I just want to make sure I'm following the right line of thought here. I was thinking a little bit about the rationals having measure zero, and how that can be pretty unintuitive if you try thinking of "measure zero" in terms of arbitrarily small open coverings; visually, if you imagine covering the rationals with arbitrarily small intervals, you're initially left with the impression that $[0,1]$ should also have measure zero, since it appears that you just covered the whole interval with an arbitrarily small open cover. So that lead me to the following.
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We show that $[0,1]$ has measure zero.
We know that $\mathbb{Q}' = \mathbb{Q}\cap[0,1]$ has measure zero.
Thus, let $\epsilon > 0$, and let $\bigcup_{i=1}^\infty (a_i, b_i)$ be an open covering of $\mathbb{Q}'$ such that $\sum_{i=1}^\infty (b_i-a_i)<\epsilon$.
Now let $x\in[0,1]$ be irrational. Since $\mathbb{Q}'$ is dense in $[0,1]$, we can find some $q\in \mathbb{Q}'$ that is arbitrarily close to $x$, and thus there is some $(a_i,b_i)$ such that $x \in (a_i,b_i)$.
It therefore follows that $[0,1] \subseteq \bigcup_{i=1}^\infty (a_i, b_i)$, and thus $\mu([0,1]) < \epsilon$.
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I think the problem with this argument is the assertion that "Since $\mathbb{Q}'$ is dense in $[0,1]$, we can find some $q\in \mathbb{Q}'$ that is arbitrarily close to $x$, and thus there is some $(a_i,b_i)$ such that $x \in (a_i,b_i)$".
It's certainly true that we can get a rational number, $q$, arbitrarily close to $x$. But what's not guaranteed is that $x$ will be contained in the part of our open covering that contains $q$. And even if we try to remedy this by picking an even closer rational number, say $q'$, we're still not guaranteed that the part of the open covering containing $q'$ will be big enough to contain $x$.
Moreover, the fact that our open covering has total length $< \epsilon$, and $\mu([0,1]) = 1$ means that there are guaranteed to be elements in $[0,1]$ that are not contained in $\bigcup_{i=1}^\infty (a_i, b_i)$, even though $\bigcup_{i=1}^\infty (a_i, b_i)$ covered a dense subset of $[0,1]$.
But we also know that the irrationals in $[0,1]$ are dense in $[0,1]$. Thus the only way you can have some irrational, $x$, not contained in your open cover is if it is the only point between (at least) two pieces of the open cover; i.e. it cannot be the case that $x$ failed to be in $\bigcup_{i=1}^\infty (a_i, b_i)$ was because there was some interval around $x$ (of any length at all) which didn't intersect with $\bigcup_{i=1}^\infty (a_i, b_i)$.
Thus, if one were to look at all the points that fail to be in our open covering, what they would see is a not-necessarily dense set of singletons which has measure at least $1-\epsilon$.
I'm sure I'm not saying anything new, and the more I think about this, the less it seems like I'm saying anything interesting at all. But the one thing that feels weird (though not wrong) is that I'm covering a dense subset of $[0,1]$ with intervals and everything that's left over is singletons; and not just any singletons, but it's actually most of them.