# Examples of no-zero-measure meagre set

I know cantor set and rational numbers in $$\mathbb{R}$$ are meagre. But they are all zero measure.

So is there any meagre set that is non-zero measure?

• What is a meagre set? Commented May 23, 2020 at 20:47
• This answer discusses how different the notions "meager set" [= first (Baire) category set] and "Lebesgue measure zero set" can be. Commented May 23, 2020 at 21:07
• @herbsteinberg A meagre set is an (at most) countable union of nowhere dense sets. A set is nowhere dense if its closure has empty interior (so is "small" topologically). Commented May 23, 2020 at 21:56

You should read about Fat Cantor sets - they are nowhere dense but have positive measure.

Yes, there is very little relationship between meagerness and measure!

The Smith-Cantor-Volterra set is an example of a meager set (in fact a nowhere dense one) with positive measure.

But the converse is also possible, one can construct a comeager set with zero measure.

The standard example (IMO) is the following construction in $$\Bbb R$$ with Lebesgue measure $$\lambda$$:

Let $$q_1, q_2,\ldots$$ be an enumeration of the rationals in $$\Bbb R$$. Let $$O(i,j) (i,j=1,2, \ldots )$$ be the open interval with centre $$q_i$$ and length $$\frac{1}{2^{i+j}}$$

Let $$U_i = \bigcup_{j=1}^\infty O(i,j)$$, and $$D=\bigcap_{i=1}^\infty U_i$$.

If $$\varepsilon>0$$ pick $$j$$ so that $$\frac{1}{2^j} < \varepsilon$$ and then note that $$D\subseteq \bigcup_{i=1}^\infty O(i,j)$$ and $$\lambda(\bigcup_{i=1}^\infty O(i,j)) \le \sum_{i=1}^\infty \lambda(O(i,j) = \sum_{i=1}^\infty \frac{1}{2^{i+j}}= \frac{1}{2^j} < \varepsilon$$

so that $$\lambda(D) = 0$$. But $$M:= \Bbb R\setminus D$$ is meagre (being the complement of an intersection of dense open sets $$U_i$$; each $$U_i$$ contains $$\Bbb Q$$ so is clearly dense) and so we can write $$\Bbb R$$ as a disjoint union of a meagre set of infinite measure and a co-meagre set of measure $$0$$.