If sets differ by a null set, they're either both measurable or both not.

Theorem: Let \((X,\mathcal M_{\mu^*}, \mu^*)\) be the measure space of Caratheodory sets induced by \(\mu^*\). Let \(E,F\subseteq X\) such that they differ by a null set, i.e. \( \mu^*(E\Delta F)=0\). Then E is measurable if and only if F is.

Proof: Clearly the symmetry of the problem allows us to prove only one direction, so assume E is measurable and we show that F is. Of course the goal is to show that \(\mu^*(A) = \mu^*(A\cap F) + \mu^*(A\smallsetminus F)\).

I will only prove that \(\mu^*(A\cap F)= \mu^*(A\cap E)\), and the proof that \(\mu^*(A\smallsetminus F) = \mu^*(A\smallsetminus E)\) will be similar. When both proofs are completed, it allows us to argue

$$ \begin{aligned} \mu^*(A) &= \mu^*(A\cap E) + \mu^*(A\smallsetminus E) \\ &= \mu^*(A\cap F) + \mu^*(A\smallsetminus F)
\end{aligned} $$

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Proof of the lemma

$$\begin{aligned} \mu^*(A\cap F) &= \mu^*(A\cap F\cap E)+\mu^*([A\cap F]\smallsetminus E) \\ &= \mu^*(A\cap F\cap E) \end{aligned}$$

where the final equality is due to the fact that \( (A\cap F)\smallsetminus E \) is a subset of the difference between E and F and therefore has measure zero.

For the next part note that every set of measure zero is measurable, and \(\mu^*(E\smallsetminus F) =0\). Therefore

$$\begin{aligned} \mu^*(A\cap E) &= \mu^*([A\cap E]\cap [E\smallsetminus F])+\mu^*([A\cap E]\smallsetminus [E\smallsetminus F]) \\ &= \mu^*([A\cap E]\smallsetminus F) + \mu^*(A\cap E\cap F)\\ &= \mu^*(A\cap E\cap F) \end{aligned}$$

Putting the two results above together, we have \(\mu^*(A\cap F) = \mu^*(A\cap E)\).