Subset of null set (set of measure zero)

Question

Is there a proper subset of a set of measure zero that is not measurable?

Any examples? Thanks a lot!

I suspect the answer is yes due to some careful phrasing in books, e.g. let F be a subset of a null set in A. If F were always measurable, it would be simpler to say let F be a null set instead.

Answer 1

The answer, in general, depends on the $\sigma$-algebra you choose.

For instance, if we take our measure space to be $(\mathbb{R}, \mathscr{A}, \lambda)$, where $\lambda$ is the Lebesgue measure, and $\mathscr{A} =\{A: \forall B, \lambda^*(B) = \lambda^*(A\cap B) + \lambda^*(A^c \cap B) \}$ is the set of all measurable (in the sense of Caratheodory's criterion) subsets, the answer is no. Clearly from the definition of $\mathscr{A}$, if $A$ is measurable and has measure $0$, and $A' \subseteq A$, then $\lambda^*(B) = \lambda^*(A' \cap B) + \lambda^*(A^c \cap B) = 0 + \lambda^*(B)$ and so $A' \in \mathscr{A}$.

However, if we take $(\mathbb{R}, \mathscr{B}(\mathbb{R}),\lambda)$ where again $\lambda$ is the Lebesgue measure, and $\mathscr{B}(\mathbb{R})$ is the borel $\sigma$-algebra on $\mathbb{R}$, then the answer is yes, though you NEED the axiom of choice in order to find one. https://mathoverflow.net/questions/32720/non-borel-sets-without-axiom-of-choice.

Answer 2

For Lebesgue Measure on $\mathbb{R}$, the answer is decisively no. The reason is Caratheodory's Theorem: Lebesgue measure is constructed as the completion of a premeasure, and this theorem guarantees that the set $\mathcal{M}$ of measurable sets under the premeasure is a $\sigma$-algebra, and the measure restricted to $\mathcal{M}$ is complete. See also Folland's Real Analysis Chapter 1.