Is this set measurable?

Let $E$ be a subset of $\mathbb{R}$.

Assume that $\forall x\in E, x$ is a limit point of $E\setminus\{x\}$.

Then, is $E$ Lebesgue-measurable?

For example, any perfect subset, open subset or connected subset are examples of $E$ that are Lebesgue measurable. I cannot think of a counterexample for this.. Is $E$ always Lebesgue measurable?

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What is the source of this question? – Did Aug 25 '14 at 10:42
Looks like the standard example at the Example part of en.wikipedia.org/wiki/Non-measurable_set is a good candidate. – Quang Hoang Aug 25 '14 at 10:54
@Did There's no source.. I'm just curious about it. – Mathemagic Aug 25 '14 at 10:56
@QuangHoang This one could have isolated points, couldn't it? – Did Aug 25 '14 at 10:59
What if you took a non-measurable subset $A$ of the irrationals and set $E=A\cup\Bbb Q$? – David Mitra Aug 25 '14 at 11:31

No, such an $E$ need not be measurable:

Let $A$ be a non-measurable set. Then $E=A\cup(\Bbb Q\setminus A)$ is non-measurable (otherwise, $A=\bigl( A\cup (\Bbb Q\setminus A)\bigr) \cap( \Bbb Q\setminus A)^C$ would be measurable).

But, as $\Bbb Q\subset E$, every point $x$ in $E$ is a limit point of $E \setminus\{x\}$.

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+1, $E$ is precisely $A\cup \Bbb Q$. – Quang Hoang Aug 25 '14 at 12:09

Lemma: Let $E \subset \Bbb{R}$ be uncountable. There is a countable set $F\subset E$ such that $E\setminus F$ satisfies your assumption.

Proof: Let

$$F := \{x \in E \mid \exists U_x \text{ neighbourhood of } x \text{ s.t. } U_x \cap E \text{ is countable}\}.$$

The open(!) covering $(U_x)_{x \in F}$ of $F$ has a countable subcover $(U_{x_n})_n$, (because $\Bbb{R}$ is 2nd countable), so that $F \subset \bigcup_n U_{x_n} \cap E$ is countable.

For $x \in E\setminus F$ and $U$ neighbourhood of $x$, we know that $E\cap U$ is uncountable. Because $F$ is countable, this yields that $E\setminus (F \cup \{x\})$ is nonempty. $\square$

Now take any nonmeasurable set $N$, choose $F\subset N$ as above. Then $N\setminus F$ is not measurable (why?), but fulfills your assumptions.

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Anything guarantees that $F$ is not less than countable? For example: $E=[0,1]$ then $F=\varnothing$. – Quang Hoang Aug 25 '14 at 12:06
It can be "less than" countable. But with my definition of countable, I mean "countably infinite or finite", so that this is the same thing. But there is no problem at all if e.g. $F = \emptyset$. – PhoemueX Aug 25 '14 at 12:59