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?
Well known proofs of this result usually do use the measurability of $A$.For example,one common way to prove the density theorem is to see it as a corollary of the Lebesgue differentiation theorem(of course,after proving it).Now here we consider the indicator function on $A$, and proceed.The key here is, if $A$ were not measurable, we wouldn't have been able to conclude that it's indicator is integrable at all(like,what would be the integral,then,when you cannot essentially measure the area under it's graph?).The notion of integrability is defined for measurable functions, for which it does make sense(to see why, you may observe that for bounded measurable functions on a closed interval, it's upper and lower Lebesgue integrals actually coincide, so we do have a consistent way to define it's integral).And coming back to your query again, Terence Tao, in his book, gives an excellent heuristic argument to demonstrate the failure of the density theorem for arbitrary sets, where no measurability is assumed.The argument he gives is, for each $x$ in $[0,1]$ consider an experiment which involves the traditional unbiased coin toss and denote it's outcome by a random variable $Y_x$ which is $1$ if you get heads and $0$ otherwise.Then, he takes $E$ to be the set of all $x$ such that $Y_x$ is $1$.He now says that in each interval, due to the number of tosses being uncountably large, $Y_x$ would equal $1$ for near about half of the points, however small the interval may be.This,in essence,is contradictory to the Lebesgue density theorem,because the density comes out to be $1$ on a set of measure near about $\frac{1}{2}$, which is far from being almost everywhere.Have a read here :-
https://terrytao.wordpress.com/2008/10/14/non-measurable-sets-via-non-standard-analysis/amp/