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?
 A: 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 $E \setminus\{x\}$.
A: 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. 
