Are these two families of sets the same without the assumption of measurability?

Question

Suppose we have a nonempty set $X$ and an outer measure $\mu$ on $\mathcal{P}(X)$. Let's denote the $\sigma$-algebra of $\mu$-measurable sets, i.e. $\sigma$-algebra from Caratheodoty extension theorem, as $S_{\mu}$. Suppose $A\subset X$. I wanted to check whether the following equality holds: $$S_{\mu|_{\mathcal{P}(A)}}=\{C\cap A:C\in S_{\mu}\}$$ I've managed to prove that $\{C\cap A:C\in S_{\mu}\}\subset S_{\mu|_{\mathcal{P}(A)}}$. Also I've managed to prove the reverse inclusion under the assumption that $A\in S_{\mu}$. Now my question is this, is the reverse inclusion true without the mentioned assumption ? I would be grateful if you could give me some hint to prove this statement or hint on constructing counterexample.

Answer 1

Let $X = \{a, b, c\}$ and $\mu(\{a\}) = \mu(\{b\}) = \mu(\{c\}) = 1$, $\mu(\{a, b\}) = \mu(\{a, c\}) = \mu(\{b, c\}) = 2$ and $\mu(\{a, b, c\}) = 10$.

Then $S_\mu = \{\varnothing, X\}$.

Take $A = \{a, b\}$. Then $S_{\mu|_{\mathcal P(A)}} = \mathcal P(A)$