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As far as I know, the Cantor set is a Borel set because it is the union of a countable collection of closed sets. Now is *any* subset of the Cantor set a Borel set?

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No.

The cardinality of the Borel $\sigma$-algebra on $\mathbb R$ (or on $[0,1]$) is the same as that of $\mathbb R$. This is a non-trivial fact that can be proved by transfinite induction; see, for example, Billingsley (1995). On the other hand, the cardinality of the set of all subsets of the Cantor set is the same as that of $2^{\mathbb R}$, the power set of $\mathbb R$—remember that the Cantor set has the cardinality of the continuum. The cardinality of $2^{\mathbb R}$, in turn, is *strictly* greater than that of $\mathbb R$ by a theorem also thanks to Cantor. Hence, at least one (vast underestimation here) subset of the Cantor set must be non-Borel.

However, any subset of the Cantor set is Lebesgue measurable by the definition of the Lebesgue $\sigma$-algebra, which is the completion of the Borel $\sigma$-algebra, given that the Cantor set has Lebesgue measure $0$.

The same reasoning shows that any subset of $\mathbb R$ that has the same cardinality as $\mathbb R$ (for instance, any non-degenerate interval) contains non-Borel subsets. In particular, there *are* non-Borel subsets of $\mathbb R$.

At any rate, note that the axiom of choice—or its equivalent variants like the well-ordering theorem invoked in transfinite induction—is crucial for all of these results. Indeed, the possibility that *all* subsets of $\mathbb R$ are Borel is logically consistent with the negation of the axiom of choice. See more about this here.

BTW, the Cantor set is itself closed, *a fortiori* Borel.

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