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I'm looking back at my book for my Real Analysis course, and the book (Royden & Fitzpatrick's Real Analysis, 4 ed.) gives the following for the definition of "almost everywhere" and the statement of the Borel-Cantelli Lemma, respectively:

(1) For a measurable set $E$, we say that a property holds almost everywhere on $E$, or it holds for almost all $x\in E$, provided there is a subset $E_0$ of $E$ for which $m(E_0)=0$ and the property holds for all $x\in E \setminus E_0$.

(2) The Borel-Cantelli Lemma $\quad$ Let $\{E_k\}_{k=1}^\infty$ be a countable collection of measurable sets for which $\sum_{k=1}^\infty m(E_k) \lt \infty$. Then almost all $x\in\mathbb{R}$ belong to at most finitely many of the $E_k$'s.

Now, my question is this: Does this imply that, for instance, almost all $x\in\mathbb{R}$ belong to the interval $[0,1]$? I have reasoned as follows:

Take $E_1 = [0,1]$ and $\{E_k\}_{k\ge2} = \emptyset$. Then $\sum_{k=1}^\infty m(E_k) = 1 + 0 + 0 + \cdots = 1 \lt \infty$. Thus the hypotheses of the Lemma are satisfied and we can conclude that almost all $x\in\mathbb{R}$ belong to $[0,1]$.

However, I struggle to find an $E_0$ that satisfies the definition of almost all. I mean that the property "belongs to $[0,1]$" holds for almost all $x\in\mathbb{R}$ by the above, so by definition, there is a subset $E_0 \subseteq \mathbb{R}$ such that $m(E_0)=0$ and "belongs to $[0,1]$" holds for all $x\in\mathbb{R}\setminus E_0$. I also find this confusing because it would seem that in order for this to be true we would have to remove at least the set $(-\infty,0)\cup(1,\infty)$ from $\mathbb{R}$ to obtain a set of numbers contained in $[0,1]$, right? But then this set doesn't have Lebesgue measure zero, right? Please let me know what I am missing here or what I am getting wrong. It seems that there is either something I am not understanding correctly or I have made a false statement somewhere.

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  • $\begingroup$ If $y$ is not in any $E_k$, then it is at most in finitely many. $\endgroup$ – Enredanrestos Mar 2 '18 at 23:55
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Almost all $x\in\Bbb R$ belong to finitely many of the $E_k$'s.

Your mistake is that you excluded zero as a 'finitely many', i.e. some $x$ may not be contained in any of the $E_k$'s.

So, in your example, actually every real number appears at most in one $E_k$, so the nullset you're after is just $\emptyset$.

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  • $\begingroup$ Thanks! I didn't expect the answer to be so succinct but I see clearly now. A great deal of my confusion came from a professor who claimed this was equivalent to the statement that (0,1) and the real line are homeomorphic, which while true, was unsatisfying and seemingly irrelevant. $\endgroup$ – dirtydivider Mar 3 '18 at 0:04
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There is no contradiction, in fact Borel-Cantelli in your example says to us that

"almost all $x\in\mathbb{R}$ belong to at most finitely many of the $E_k$'s"

so could be exists $x \in \mathbb{R}$ such that $x \not \in E_k \forall k$ without any contradiction.

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