If $A subset mathbb{R}^{n}$ is Lebesgue measurable, then we see that $x in A$ is a point of density if

$$

lim_{r to 0} frac{mu(A cap B(x,r))}{mu(B(x,r))} =1

$$

The Lebesgue Density Theorem says that almost every point of $A$ is a point of density. Can I extend this notion of density to using the outer measure (for any subset of $mathbb{R}^{n}$) which I, a priori, don’t know is measurable? Does the Lebesgue Theorem apply in this case? My guess is yes, because the proof doesn’t really use the fact that $A$ is measurable. I want to use density to show that a set is measurable.

Tl;dr – obviously for measurable sets it makes no difference if I use “measure” or “outer measure”, but is there a notion of density for non-measurable sets (or sets not a priori measurable) using outer measure?