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Let $r\in[0,1]$. We look at the binary represenation of $r$ and say that $r$ is binarily universal if every finite binary string appears in at least one place in the binary representation of $r$. Let $U$ be the set of binarily universal members of $[0,1]$.

Is $U$ a Borel set? If yes, what is its Borel measure?

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    $\begingroup$ For a finite binary string $b$, the set $A_b$ of those $r \in [0, 1]$ which do not contain $b$ as a substring in the binary representation of $b$ is clearly Borel. There are countably many strings $b$, so the union of $A_b$ is also Borel. The set $U$ is the complement of this union. The Lebesgue measure of $A_b$ is zero, so $U$ is of full Lebesgue measure. (Actually, $U$ contains the set of normal numbers, which is of full Lebesgue measure). Does this answer your question? $\endgroup$
    – Mateusz Kwaśnicki
    Mar 13, 2018 at 8:48

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$U$ is a $G_\delta $ set, hence Borel at level 2 of the Borel hierarchy. It has Lebesgue measure 1 and is also comeager, so it is large both in the sense of measure and of Baire category.

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