I'd like to prove the last part of an exercise in my book. The quetion is as follows
Let $(X,\Sigma ,\mu )$ be a measure space and define $$\mu '(M):=\inf\left\{ \mu (A):M\subset A\in \Sigma \right\} $$ as the outer measure ( I already showed that this is an outer measure). $M\in { 2 }^{ X }\setminus \Sigma $ and $\mu'(M)>0$ further let $\Sigma_M$ be the sigma algebra generated by $M$. Find all measures ${ \mu }_{ M }$ on ${ \Sigma }_{ M }$ with ${ { \mu }_{ M } }_{ |\Sigma }={ \mu }$.
My first question would be if there is a typo in my book shouldn't it be ${ { \mu }_{ M } }_{ |\Sigma }={ \mu' }$?
My second question is how do I construct such measures i can't even think of possible canidates. Could someone provide a hint? I'm lost on this one.
Firstly, we already have a measure $\mu$ on $\Sigma$ so it is actually logical to ask for measures on the larger $\sigma$-algebra $\Sigma_M$ (which IMHO should be desribed as the $\sigma$-algebra generated by $\Sigma \cup \{M\}$, not as a trace or some such notion; just the "$\sigma$-algebra generated by $M$" makes no sense in this context: that is just the $4$-set $\sigma$-algebra $\{\emptyset, X, M, X\setminus M\}$ and is unrelated to $\Sigma$), that are extensions of $\mu$. So $(\mu_M)_{|_{\Sigma}} = \mu$ seems the right condition to me.
The prime candidate for such a $\mu_M$ is of course the given $\mu'$ which coincides with $\mu$ on $\Sigma$ already (for $M \in \Sigma$ the infimum described is a minimum attained at $M$ itself..), but then we get into Carathéodory theorem realms, whether this $\mu'$ is actually a well-defined measure on $\Sigma_M$, look up theorems around that.