Outer Lebesgue measure of a subset is equal to the measure of the whole set

Question

When $A\subset X$ has $\nu(A)=\mu(X)<\infty$ where $\nu$ is the outer Lebesgue measure and $\mu$ is the Lebesgue measure, can I conclude that $A$ is dense?

I think $\nu(\bar{A})=\mu(\bar{A})=\mu(X)$ but don't know where to get from here.

Answer 1

If $A$ was not dense, then there is some open (hence Lebesgue measurable) ball $B$ such that $B \cap A = \emptyset$. In particular, $A \subset X \setminus B$, and hence $\nu A \le \mu X - \mu B < \mu X$.