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?
Best Answer
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 $0$.