Weakest condition for a subset of $\mathbb{R}$ to have positive Lebesgue measure I just started studying measure theory and I'm interested to know what's the weakest criterion used by mathematicians in order to determine whether a subset of $\mathbb{R}$ has positive Lebesgue measure. 
Initially I thought that it would be sufficient that the closure of this set didn't have empty interior. However, I recently found examples of a nowhere dense set that has positive Lebesgue measure, the Smith-Volterra-Cantor set. 
 A: [WARNING this answer is not correct. The starting train of thought is fine but in the end one cannot conclude what I concluded. For pedagogical purposes and because of the value of the comments, I won't delete the answer.]
Let $S $ be a subset of $\mathbb{R} $. Its Lebesgue measure is given by the Lebesgue integral
$$\int_{-\infty}^\infty \chi_S(x) dx $$
Where $\chi_S $ is the characteristic function of $S $. Since $\chi_S $ is always 0 or 1, the integral above is always, at least, 0. That is true even if we change the bounds. For example if we knew $S \subset [a, b] $ then its measure is
$$\int_{a}^b \chi_S(x) dx $$ since $\int_{-\infty}^a\chi_S(x) dx = \int_{b}^\infty \chi_S(x) dx = 0$
Therefore if $$\int_{-\infty}^\infty \chi_S(x) dx > 0$$ there exist $b > a $ such that
$$\int_{a}^b \chi_S(x) dx > 0 $$
INCORRECT STARTING FROM HERE
which is the same as saying that $S $ contains an interval $[a, b] $ where $\chi_S $ is not 0 almost everywhere. Which is the same as saying that $\chi_S $ is 1 almost everywhere which means $\chi_S = 1$ in $[a, b] $ for an uncountable number of points.
EXPLANATION OF THE ERROR
Saying $\chi_S $ is 1 almost everywhere in $[a, b] $ is equivalent to saying $S $ is dense in $[a, b] $ which need not be true nor sufficient for $S $ to have positive measure, as one can see from two examples: the rationals are dense in every interval and have measure 0. Cantor sets are dense nowhere and have positive measure.
