Examples of no-zero-measure meagre set

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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 $0$.