# Are all sets of zero measure (not necessarily Lebesgue measure) measurable?

I know that all sets with Lebesgue outer measure equal to zero are measurable. This is not hard to prove using the definition of a Lebesgue measurable set, E $$\subseteq X$$, that I was given : $$m^*(A)= m^*(A\cup E) + m^*(A\cup E^C) ,\forall A\subseteq X$$

But what about the case in a general measure space, $$(X,\mathfrak{X} ,\mu)$$?

Say $$\mu (E)= 0$$. Does this imply that $$E$$ is measurable?

All I have about general measurable sets is that they belong to $$\mathfrak{X}$$, where $$\mathfrak{X}$$ is a subset of the power set of $$X$$ that is closed under compliments and unions and contains $$X$$ and $$\emptyset$$. All I have about the measure, $$\mu$$, is that it maps from $$\mathfrak{X}$$ to $$[0,\infty]$$, $$\mu(\emptyset)=0$$ and we have disjoint additivity.

Does the very fact that $$\mu(E)$$ is defined imply that E is measurable? This would seem, perhaps, obvious but I am not confident with Measure Theory. I find it best not to make assumptions.

• $\mu$ is only defined on measurable sets. So writing $\mu(E)=0$, implicitly assumes that $E$ is measurable. Commented Nov 30, 2020 at 11:15
• Thank you for verifying that for me. It is what I assumed but got thrown off by the concept of the Lebesgue outer measure being defined for all sets in $\mathbb{R}$ whether or not they are measurable. Can I make your comment the official answer to my question? Commented Nov 30, 2020 at 11:22

Well, $$\mu$$ is a map $$\mathfrak X\longrightarrow[0,\infty]$$, so the fact that $$\mu(E)$$ is defined in the first place makes $$E$$ an element of $$\mathfrak X$$, so measurable.
If you're asking about outer measures (which measure spaces don't need!), then your question makes more sense and the answer is still yes, sets of measure $$0$$ are measurable. Let $$\mu^\ast:\mathcal P(X)\longrightarrow [0,\infty]$$ be an outer measure, $$E\subseteq X$$ such that $$\mu^\ast(E)=0$$. $$E$$ is measurable by definition if for all $$A\subseteq X$$ we have $$\mu^\ast(A)=\mu^\ast(E\cap A)+\mu^\ast(E^c\cap A)$$. Since outer measures are $$\sigma$$-subadditive this is equivalent to $$\mu^\ast(A)\geq \mu^\ast(E\cap A)+\mu^\ast(E^c\cap A)$$. Because of monotonicity we have $$\mu^\ast(E\cap A)\leq\mu^\ast(E)=0$$, so $$\mu^\ast(E\cap A)=0$$. So we only need to show $$\mu^\ast(A)\geq \mu^\ast(E^c\cap A)$$. But that's just monotonicity of the outer measure. So $$E$$ is measurable.
• Just want to check if I understood correctly. Are you saying, by $\sigma$-subadditivity of outer measures, since $A\subset \left(E\cap A\right) \cup \left(E^c \cap A\right) \cup_i \emptyset$, we have $\mu^*(A)\leq \mu^*\left(E\cap A\right) + \mu^*\left(E^c \cap A\right)+0+\cdots$, and so we only need to prove the $\geq''$ part? Commented Nov 22, 2022 at 14:08