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.

• Using these "fat Cantor sets" you can actually arrange a meager set with a full measure in the unit interval, whose complement is therefore a null set which is co-meager. This shows that the interaction between [Lebesgue] measure and [Baire] category is more or less orthogonal when it comes to arbitrary sets. – Asaf Karagila Nov 12 '16 at 11:54
• Would in it be to contain an interval of the form $[a, b]$? All intervals of the form $[a, b], a \not= b$ have positive lebesgue measure so that should do. – RGS Nov 12 '16 at 11:57
• @RSerrao That's actually a strong condition since we have nowhere dense sets with positive measure. – user93511 Nov 12 '16 at 11:58
• @AsafKaragila I know nothing about Baire categories. But, I'll look into this topic. – user93511 Nov 12 '16 at 11:59
• Baire category is the measure-analogue for general topological spaces, more-or-less. Specifically, a set is meager if it is the countable union of closed sets with empty interior. So a co-meager set is the countable intersection of dense open sets. In complete metric spaces, it turns out that a co-meager set is also dense. And so it turns out to be a good measure for "being large" (note that just being dense is not enough, as there are countable dense sets in $\Bbb R$). But the closure having an empty interior is exactly the first step into the Baire category territories. – Asaf Karagila Nov 12 '16 at 12:07

[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.

• Well. That's a long way of saying "A measurable set is non-null, if it has a positive measure". – Asaf Karagila Nov 12 '16 at 12:37
• @AsafKaragila not really. I showed something a bit more useful than that. If $S$ has positive measure then there is an interval such that the intersection of $S$ with that interval is dense on that interval. And that is both necessary and sufficient. – RGS Nov 12 '16 at 13:38
• If $S$ is a fat cantor set then it is compact, has positive measure and is not dense in any interval of positive measure (since it is compact and totaly disconnected). And since rationals are dense but of measure $0$ it's neither necessary nor sufficient. – Renart Nov 12 '16 at 13:38
• @Renart right... then I am not being able to express what I mean with the right set properties. If the characteristic function is 1 almost everywhere inside [a,b] then what does it say about $[a,b] \cap S$? That it is uncountable? Can't I get anything more strong than that? – RGS Nov 12 '16 at 13:46
• @RSerrao The point of Renant's comments, along with similar comments on the OP, is that there is essentially no weak-yet-sufficient topological condition for a set to have positive measure. There is the strong topological condition of having a nonempty interior...and that's pretty much it, because even the "topologically smallest" sets (i.e. nowhere dense sets) can have positive measure. – Ian Nov 12 '16 at 13:54