Show that $m(\mathbb{A})=0$ where $\mathbb{A}=\{x\in \mathbb{B} \mid \exists \delta >0 \quad m\left((x-\delta,x+\delta)\cap \mathbb{B} \right) =0 \}$ Let $\mathbb{B} \subset \mathbb{R}$ is a measurable set and define
$$\mathbb{A}=\{x\in \mathbb{B} \mid \exists \delta >0 \quad m\left((x-\delta,x+\delta)\cap \mathbb{B} \right) =0  \}.$$
Show that $m(\mathbb{A})=0$.
Since  $\mathbb{A} \subset \mathbb{B}$,
$$m(\mathbb{A})=m(\mathbb{A} \cap \mathbb{B})\leq m(\cup_{x\in \mathbb{A} }(x-\delta_0,x+\delta_0)\cap \mathbb{B}) \leq \sum_{x\in \mathbb{A} } m\left((x-\delta_0,x+\delta_0)\cap \mathbb{B} \right)=0,$$
where $\delta_0$ is minimum of all $\delta_x$. Is this solution valid?
 A: Let $\{x_n\}$ be a countable sequence enumerating the rational numbers. The set of intervals $(x_n - 1/m, x_n +1/m)$ is countable as the cartesian product of two countable sets is countable. Let $\{I_n\}$ be a sequence enumerating those intervals.
Now let
$$\mathbb A^\prime = \{x \in \mathbb B \mid \exists n \in \mathbb N (x \in I_n \text{ and } \ m(I_n \cap \mathbb B) = 0)\}.$$
I claim that $\mathbb A = \mathbb A^\prime$. The inclusion $\mathbb A^\prime \subseteq \mathbb A$ is clear. Conversely, for $x \in \mathbb A$, it exists $\delta \gt 0$ with $m\left((x-\delta,x+\delta)\cap \mathbb{B} \right) =0$. Let $m \in \mathbb N$ be such that $1/m \lt \delta/2$ and $n \in \mathbb N$ such that $\vert x - x_n \vert \lt 1/m$. For $y \in (x_n-1/m, x_n +1/m)$ we have
$$\vert x- y \vert \le \vert x - x_n \vert + \vert x_n - y \vert \le 2/m \lt \delta$$ proving that $(x_n-1/m, x_n +1/m) \subseteq (x-\delta,x+\delta)$ and therefore that $x \in \mathbb A^\prime$.
From there, you get that $\mathbb A$ is included in a countable union of null sets and is, therefore, a null set.
