The complement of a measurable cover is the measurable cover of the complement? Let $(\Omega, A, P)$ be a probability space, $X: \Omega \rightarrow S$ a non-measurable mapping in the metric space $S$.
The sets $(X \in C)$ with $C$ a subset of $S$ are not measurable.
We define the measurable covers of such sets as the smallest (in terms of inclusion) measurable sets that cover $(X \in C)$, we use the notation $(X \in C)^*$.
Supposing that such set exists for a fixed $C$, is it true that $((X \in C)^*)^c = (X \not \in C)^*$
Same question if we define $(X \in C)_*$ as the biggest (in terms of inclusion) measurable set included in $(X \in C)$.
I feel like there is something to do with inner and outer probabilities, but I can't see how.
 A: By a measurable cover $\{X \in C\}^*$ of $\{X \in C\}$, I believe you mean that $\{X \in C\}^*$ is measurable, $\{X \in C\}^* \supseteq \{X \in C\}$, and for all other measurable $A$ with $A \supseteq \{X \in C\}$ we have $\{X \in C\}^* \subseteq A$. 
Similarly, $\{X \in C\}_*$ is (if it exists) measurable with $\{X \in C\}_* \subseteq \{X \in C\}$, and for all measurable $A$ with $A \subseteq \{X \in C\}$ we have $\{X \in C\}^* \supseteq A$.
Claim. If $\{X \in C\}^*$ exists, then so does $\{X \notin C\}_*$ and $(\{X \in C\}^*)^c = \{ X \notin C\}_*$.
Proof. It suffices to show that $(\{X \in C\}^*)^c$ is a measurable subset of $\{X \notin C\}$ with the property that if $A$ is a measurable subset of $\{X \notin C\}$, then $ A \subseteq (\{X \in C\}^*)^c$. 
Since $\{X \in C\}^*$ is measurable, so is $(\{X \in C\}^*)^c$.
Since $\{X \in C\}^* \supseteq \{X \in C\}$, then $(\{X \in C\}^*)^c \subseteq \{X \notin C\}$.
If $A$ is measurable and $A \subseteq \{X \notin C\}$, then $A^c$ is measurable and $A^c \supseteq \{X \in C\}$. Hence, $A^c \supseteq \{X \in C\}^*$, which implies $A \subseteq (\{X \in C\}^*)^c$.
QED.
