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A mathematician

One might say, "a random subset of $\mathbb{R}$ is not Lebesgue measurable" without really thinking about it. But if we unpack the standard definitions of all those terms (and work in ZFC), it's not so clear.

Let $\Sigma \subset 2^\mathbb{R}$ be the sigma-algebra of all Lebesgue measurable sets. Give $2^\mathbb{R}$ the product measure. (It's a product of continuum many copies of the two-point set.) We want to say that $\Sigma$ is a null set in $2^\mathbb{R}$...but is $\Sigma$ even measurable?

Laci Babai posed this question casually several years ago, and no one present knew how to go about it, but it might be easy for a set theorist.

Also, a related question: Think of $2^\mathbb{R}$ as a vector space over the field with two elements and $\Sigma$ as a subspace. (Addition is xor, that is, symmetric set difference.) What is $\dim\left(2^\mathbb{R}/\Sigma\right)$?

It's not hard to see that $\dim\left(2^\mathbb{R}/\Sigma\right)$ is at least countable, so if $\Sigma$ were measurable, it would be a null set. But that's as far as I made it.

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# Is a random subset of the real numbers non-measurable? Is the set of measurable sets measurable?

A mathematician might say, "a random subset of $\mathbb{R}$ is not Lebesgue measurable" without really thinking about it. But if we unpack the standard definitions of all those terms (and work in ZFC), it's not so clear.

Let $\Sigma \subset 2^\mathbb{R}$ be the sigma-algebra of all Lebesgue measurable sets. Give $2^\mathbb{R}$ the product measure. (It's a product of continuum many copies of the two-point set.) We want to say that $\Sigma$ is a null set in $2^\mathbb{R}$...but is $\Sigma$ even measurable?

Laci Babai posed this question casually several years ago, and no one present knew how to go about it, but it might be easy for a set theorist.

Also, a related question: Think of $2^\mathbb{R}$ as a vector space over the field with two elements and $\Sigma$ as a subspace. (Addition is xor, that is, symmetric set difference.) What is $\dim\left(2^\mathbb{R}/\Sigma\right)$?

It's not hard to see that $\dim\left(2^\mathbb{R}/\Sigma\right)$ is at least countable, so if $\Sigma$ were measurable, it would be a null set. But that's as far as I made it.