Extension of measures

Question

Here an exercise of the book: A probability path by Sidney Resnick.

Suppose $P$ is a probability in a $\sigma$-field $\cal{B}$ and $A\notin \cal{B}$. Let $\mathcal{B}_1:=\sigma(\mathcal{B}\cup\{A\})$ the sigma field generated by $\cal{B}$ and $A$. Show that $P$ has an extension to a probability measure $P_1$ in $\cal{B}_1$.

My problem: I can not see how to use the extension theorem of Caratheodory. My first attempt was the following, the smallest or-field containing both $\cal{B}$ and $A$ consists of sets of the form $$ C=(A\cap B)\cup (A^{c}\cap B^{'})~~B,B^{'}\in \mathcal{B}. $$ Then we define a probability P on sets above the obvious way: $$ P_1(C)= \inf \{\sum P(B_i);~~ C\subset \bigcup_{i=1}^{\infty} B_i,~~B_i\in \mathcal{B}\} $$ where the infimum is taken over all covers of C by elements of the sigma field $\mathcal{B}$. My strategy and try to reduce the problem to the application of the some know extension theorem. This is the right way?

Answer 1

Elaborating on the answer: Without loss of generality, we can assume that an inner measure of $A$ is zero(i.e. $P_*(A)=0$) and outer measure is positive(i.e., $P^*(A)>0$). Let $A_1$ be such measurable set that $A\subset A_1$ and $P(A_1)=P^*(A)$. For $C=(A\cap X)\cup (A^c \cap Y)$($X,Y \in \cal{B}$) we set $P_1(C)=P(A_1\cap X)+P(((A_1)^c) \cap Y)$. Then $P_1$ is an extension of $P$ defined on $\cal{B}_1$.

• If $P(A)$ is not zero, let $X$ be the measurable kernel of $A$ and replace $A$ by $A\backslash X$. The $\sigma$-algebra $\mathcal{B}_{1}$ is generated by $\mathcal{B}$ and $A\backslash X$, and $P_{*}(A\backslash X)=0$, so we can replace $A$ by $A\backslash X$ and assume $P_{*}(A)=0$.
• The set $A_{1}$ is a measurable envelope of $A$ which must exist.
• Every set in $\mathcal{B}_{1}$ has the form $(A\cap X)\cup (A^{c}\cap Y)$ for some sets $X,Y\in\mathcal{B}$. (This is problem 34 of section 1.9 of Redneck's A Probability Path, and to prove it one shows that the indicated sets belong to $\mathcal{B}_{1}$ and that this class of sets is in fact a $\sigma$-algebra).
• We need to show that $P_{1}$ as above is well-defined. For this we need to check that if $X\cap A=X'\cap A$ then $P(X\cap A_{1})=P(X'\cap A_{1})$ with a similar statement for $Y$. Now $P(X\cap A_{1})=P((X\cap X')\cap A_{1})+P((X\backslash (X\cap X'))\cap A_{1})$. However, $Y(X\backslash (X\cap X'))\cap A_1$ satisfies $Y\cap A=\emptyset$ since $X\cap A=X'\cap A$. Since $A_{1}$ is a measurable envelope for $A$, and $Y$ is measurable, we must have $P(A_{1}\backslash Y)=P(A_{1})$ so $P(Y)=0$. So $P(X\cap A_{1})=P((X\cap X')\cap A_{1})=P(X'\cap A_{1})$ by a symmetric argument.
• An analogous argument works for $Y$.

Answer 2

Without loss of generality, we can assume that an inner measure of $A$ is zero(i.e. $P_*(A)=0$) and outer measure is positive(i.e., $P^*(A)>0$). Let $A_1$ be such measurable set that $A\subset A_1$ and $P(A_1)=P^*(A)$. For $C=(A\cap X)\cup (A^c \cap Y)$($X,Y \in \cal{B}$) we set $P_1(C)=P(A_1\cap X)+P(((A_1)^c) \cap Y)$. Then $P_1$ is an extension of $P$ defined on $\cal{B}_1$.