# How “compact” are sets of finite measure?

Let $K$ be a compact set of $\mathbb R^n$, then every open cover of $K$ will have a finite subcover.

Now consider the following situation: Everything I say in the following is with respect to the standard Borel sigma algebra with Lebesgue measure.

Let $M$ be a bounded measurable subset of $\mathbb R.$ For every $x \in M$ let there be an $\varepsilon_x>0.$ In the sequel we write $M_x^{\alpha}:=(-\alpha+x,\alpha+x)$.

The following "(almost)-compactness" up to arbitrary small measure is easy shown to be true: Let $\varepsilon>0$ be given. There are finitely many sets $M_x^{\varepsilon_x}$ that cover $M$ up to a set of measure $\varepsilon.$ This is just by the continuity of the Lebesgue measure.

What is less obvious to me is whether the same property holds, if we want to make the sets disjoint:

Are there finitely many sets $M_{x_i}^{\delta_{x_i}}$ with $x_i \in M$ and $\delta_{x_i}\le \varepsilon_{x_i}$ such that those sets $M_{x_i}^{\delta_{x_i}}$ do not overlap and cover $M$ up to a set of arbitrary small measure?

More precisely the question is: For given $\varepsilon>0$ is there a finite number of such disjoint sets $M_{x_i}^{\delta_{x_i}}$ such that $M\backslash \bigcup_{i=1}^n M_{x_i}^{\delta_{x_i}}$ is a set of measure at most $\varepsilon$. Also, the $M_{x_i}^{\delta_{x_i}}$ do not have to be contained in $M,$ as this is not possible in general.

Here we would need an additional shrinking step, to make the sets disjoint. So after I pick the first set, I have to shrink all the remaining sets such that they are disjoint from my first one and then start looking among those, again. I do not fully understand how to do it in this case and would like to know whether this is still true or generally impossible?

Vitali's covering theorem says if you take a sequence of balls $M_{x_i}^{\epsilon_i}$ such that every $x \in M$ is covered by balls of arbitrarily small diameter, then there is a disjoint subsequence whose union contains $M$ except for a set of measure $0$.