Is a subset of [0, 1] whose points are "isolated in measure" necessarily null?

Question

Suppose that $E \subseteq [0, 1]$ has the property that for each $x \in E$, there is $r_x > 0$ such that $B(x, r_x) \cap E = (x-r_x, x+r_x) \cap E$ has Lebesgue measure zero. Is it necessarily the case that $E$ itself has Lebesgue measure zero?

A couple remarks:

1) If the points of $E$ were legitimately isolated from one another, $E$ would necessarily be countable, and hence have measure zero.

2) Because I am only interested in the question of "measure zero or not," measurability does not appear explicitly in the statement of the question above. And indeed, to answer the question in the negative, one does not need to produce a set of positive measure; producing a Lebesgue non-measurable set $E$ with the given property would just as well constitute a negative answer.

Answer 1

$E$ is Lindelöf, as a separable (second countable) metric space.

So we have a cover of $E$ by these $(x-r_x,x+r_x)$ and so a countable subcover $\{(x-r_x,x+r_x)\cap E: x \in N \}$ exists for some countable subset $N$ of $E$.

But then (as we have a cover of $E$): $$\lambda(E) \le \sum_{x \in N} \lambda((x-r_x,x+r_x) \cap E) = 0$$

by countable subadditivity of the Lebesgue measure $\lambda$. So $\lambda(E)=0$ indeed.

Answer 2

Yes, it is true. Let $U=\cup_{e\in E}B(e,r_e)$. As $U$ is an open subset of $[0,1]$, it is a countable disjoint union of intervals, say $U=\cup_{n\in\mathbb N}I_n$ with $I_n, I_m$ disjoint for $n\not=m$. Then one has $$ E=\bigcup_{n\in\mathbb N}E\cap I_n $$ where each set $E\cap I_n$ is null, and hence so is $E$.