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.
Best Answer
The answer to your second question (assuming the axiom of choice, to dodge Asaf's comment) is that $2^{\mathbb R}/\Sigma$ has dimension $2^{\mathfrak c}$, where $\mathfrak c=2^{\aleph_0}$ is the cardinality of the continuum. The main ingredient of the proof is a partition of $[0,1]$ into $\mathfrak c$ subsets, each of which intersects every uncountable closed subset of $[0,1]$. To get such a partition, first note that there are only $\mathfrak c$ closed subsets of $[0,1]$, so you can list them in a sequence of length (the initial ordinal of cardinality) $\mathfrak c$ in such a way that each closed set is listed $\mathfrak c$ times. Second, recall that every uncountable closed subset of $[0,1]$ has cardinality $\mathfrak c$. Finally, do a transfinite inductive construction of $\mathfrak c$ sets in $\mathfrak c$ steps as follows: At any step, if the closed set at that position in your list is $C$ and if this is its $\alpha$-th occurrence in the list, then put an element of $C$ into the $\alpha$-th of the sets under construction, being careful to use an element of $C$ that hasn't already been put into another of the sets under construction. You can be this careful, because fewer than $\mathfrak c$ points have been put into any of your sets in the fewer than $\mathfrak c$ preceding stages, while $C$ has $\mathfrak c$ points to choose from. At the end, if some points in $[0,1]$ remain unassigned to any of the sets under construction, put them into some of these sets arbitrarily, to get a partition of $[0,1]$.
Once you have this partition, notice that every piece has outer measure 1, because otherwise it would be disjoint from some closed set that has positive measure and is therefore uncountable. This implies that, among the $2^{\mathfrak c}$ sets that you can form as unions of your partition's pieces, only $\varnothing$ and $[0,1]$ can be measurable. In particular, no finite, nonempty, symmetric difference of these pieces is measurable. That is, they represent linearly independent elements of $2^{\mathbb R}/\Sigma$.