$\lim_{k \to \infty}v(E_k) = v(E)$

Question

$E_1,E_2, \dots$ are measurable sets, $E \subset E_i$ for all $i$ is also a measurable set and bounded.

In addition we are given that for all $k \in \mathbb N$ and for all $x \in E_k$ there is a $y \in E$ such that $|x-y| < \frac{1}{k}$

We are asked to prove that $\lim_{k \to \infty}v(E_k) = v(E)$, where $v$ stands for volume.

I understand the idea, $E_k$ gets closer and closer to $E$, until the maximum distance between $E_k$ and $E$ becomes infinitesimally small, but I don't know how to write a formal proof for this idea.

We are working with jordan measure here (as in, boundary is negligible)

Answer 1

I believe I may have found a solution.

I will attempt to prove that $\lim_{k \to \infty}v(E_k\setminus E) = 0$ which is identical.

(*) Consider $x \in E_k \setminus E$. The closest point $y \in E$ to it can't be an interior point, so infact $y \in \partial E$ and $|x-y| < \frac{1}{k}$.

Let $\epsilon > 0$, and $k$ such that $(\frac{2}{k})^n < \epsilon$.

Since $\partial E$ is negligible, we can cover it by countably many boxes with sum of volumes less than epsilon. Cover $\partial E$ by boxes with side length $\frac{2}{k}$.

This is also a cover of $E_k\setminus E$ because of (*), and each box has volume $(\frac{2}{k})^n < \epsilon$, so each box is negligible, and so we actually found a countable union of negligible boxes that cover $E_k \setminus E$, so it's negligible, which means it has zero volume.

Answer 2

Let $E^{(k)}$ denote $\{x:|x-y| <\frac 1 k \text {for some}\, y \in E\}$. Since the boundary of $E$ has measure $0$ it follows that $\nu (E^{(k)}) \to \nu (E)$. [Indeed, $E^{(k)}$'s are decreasing and $\cap_k E^{(k)}$ lies between $E$ and $\overline {E}$]. Since $E_k \subset E^{(k)}$ we get $\nu (E_k) \leq \nu( E^{(k)})$ so $\lim \sup \nu (E_k) \leq \nu(E)$. Since $E \subset E^{(k)}$ for all $k$ we are done.