A measure space $(X,\mathcal{S},\mu)$ can be extended to a measure space $(X,\hat{\mathcal{S}},\hat{\mu})$ as follows: $\mu$ is extended to an outer measure $\mu^\ast$ (defined as the infimum of the sums of $\mu$-measures of sets in $\mathcal{S}$ covering a subset of $X$), $\hat{\mathcal{S}}\mathrel{\mathop:}=\mathcal{M}(\mu^\ast)$ is the set of $\mu^\ast$-measurable sets, and $\hat{\mu}\mathrel{\mathop:}=\mu^\ast|_{\mathcal{M}(\mu^\ast)}$ is the restriction of $\mu^\ast$ to $\mathcal{M}(\mu^\ast)$. (Apparently the standard name for this is Carathéodory's extension.)
Now, if there is an inclusion of measure spaces $$(X,\mathcal{S},\mu)\subseteq(X,\mathcal{T},\nu),$$ i.e., $\mathcal{S}\subseteq\mathcal{T}$ and $\nu|_{\mathcal{S}}=\mu$, then does it follow that $$(X,\hat{\mathcal{S}},\hat{\mu})\subseteq(X,\hat{\mathcal{T}},\hat{\nu}) \ ?$$
I'm asking this because the completion of measure space preserves inclusion, and I wondered if the same holds for Carathéodory's extension. I tried to prove it, but unlike completion, it seems highly nontrivial.
Carathéodory's extension of measure spaces does NOT preserve inclusion, as it can be seen in the following counter-example:
Let $X= \Bbb R$, $\mathcal{S}=\{\emptyset, \Bbb R \}$ and $\mu$ be defined by $\mu(\emptyset)=0$ and $\mu(\Bbb R)=+\infty$. Clearly $(\Bbb R,\mathcal{S},\mu)$ is a measure space. Let $\mathcal{T}$ be the Borel $\sigma$-algebra in $\Bbb R$ and $\nu$ be the Lebesgue measure. Clearly, $(\Bbb R,\mathcal{T},\nu)$ is a measure space. It is also clear that $(\Bbb R,\mathcal{S},\mu)\subseteq(\Bbb R,\mathcal{T},\nu)$ (in the sense that $\mathcal{S}\subseteq\mathcal{T}$ and $\nu|_{\mathcal{S}}=\mu$).
Now, considering $(\Bbb R,\mathcal{S},\mu)$, we have that for all $A \subseteq \Bbb R$, $\mu^*(A)=0$ if and only if $A=\emptyset$ and $\mu^*(A)=+\infty$ if and only if $A \neq\emptyset$. It follows immediately that all subsets of $\Bbb R$ are $\mu^*$-measurable, that is $\hat{\mathcal{S}}=\mathcal{M}(\mu^*)= 2^{\Bbb R}$. Then $\hat{\mu}=\mu^*|_{\mathcal{M}(\mu^*)}$ is such that $\hat{\mu}(A)=0$ if and only if $A=\emptyset$ and $\hat{\mu}=+\infty$ if and only if $A \neq\emptyset$.
On the other hand, considering $(\Bbb R,\mathcal{T},\nu)$, we have that $\hat{\mathcal{T}}$ is the Lebesgue $\sigma$-algebra in $\Bbb R$ and $\hat{\nu}$ is the Lebesgue measure.
So we have that $\hat{\mathcal{S}} \nsubseteq \hat{\mathcal{T}}$ (In fact $\hat{\mathcal{T}} \subset \hat{\mathcal{S}}=2^{\Bbb R}$). Moreover, it is trivial to see that there are $E \in \hat{\mathcal{T}}$ where $\hat{\nu}(E) < \hat{\mu}(E)=+\infty$.
Remark: Note that the measure space $(\Bbb R,\mathcal{S},\mu)$ above is complete, that is, it is its own completion. So $(\Bbb R,\mathcal{S},\mu)$ is also an example where the Carathéodory's extension of a measure space is strictly larger than the completion of the measure space.