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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?

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  • $\begingroup$ What is a meagre set? $\endgroup$ Commented May 23, 2020 at 20:47
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    $\begingroup$ This answer discusses how different the notions "meager set" [= first (Baire) category set] and "Lebesgue measure zero set" can be. $\endgroup$ Commented May 23, 2020 at 21:07
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    $\begingroup$ @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). $\endgroup$ Commented May 23, 2020 at 21:56

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

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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.

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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 $0$.

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