Does the complement of every dense open subset of the real numbers have Lebesgue measure $0$?
This is certainly not a characterization of dense open subsets of reals, since the complement of the irrationals(which is not open) has Lebesgue measure $0$. So, is there any useful characterization of dense open subsets of the reals? Also, please mention what happens if the Axiom of Choice is not assumed.
No, it does not. Fix an enumeration of the rationals $\{r_k\}_{k = 0}^{\infty}$ and put an open interval
$$\mathcal{O}_k = \left(r_k - \frac{1}{2^{k + 1}}, r_k + \frac{1}{2^{k + 1}}\right)$$
around this point. Define $$\mathcal{O} = \bigcup_k \mathcal{O}_k$$
in which case $\mathcal{O}$ is an open, dense subset of $\mathbb{R}$ with Lebesgue measure at most $2$ (and by lengthening any one of the intervals, we can make it exactly $2$); hence its complement has infinite measure (and as far as I see, choice has nothing to do with this).
This can be modified easily so that the complement has arbitrarily large or small (finite) measure. I don't really see that there's a useful (measure-theoretic) characterization beyond the definition.