Exercise 1.7.7 (Hahn-Kolmogorov extension is unique up to a restriction)

This exercise is closely related to Exercise 1.4.28 (Approximation by an algebra).

We use the usual trick of demonstrating that the left-hand side is both less than or equal to the right-hand side and vice versa.

• $\forall <!--\; FUNCTION\; APPLICATION\; -->E\in B\cap B^{\prime }:{\mu }^{\prime }(E)\le \mu (E)$.
  This assertion can be viewed as an implication of

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->B^{\prime }\in B^{\prime }:{\mu }^{\prime }(B^{\prime })\le {\mu }^{\text{∗}}(B^{\prime }).$$

  Making use of the fact that ${\mu }^{\prime }={\mu }_0$ when restricted to $B_0$ and ${\mu }^{\prime }$ is countably additive, we can rewrite ${\mu }^{\text{∗}}(B^{\prime })$ as follows:

  $${\mu }^{\text{∗}}(B^{\prime })=\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{\sum _{n=1}^{\infty }{\mu }_0(E_n)|\begin{array}{c}{B}^{\prime }\subseteq \bigcup _{n=1}^{\infty }{E}_n\hfill \\ E_n\in B_0\hfill \end{array}\right\}=\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{\sum _{n=1}^{\infty }{\mu }^{\prime }(E_n)|\begin{array}{c}{B}^{\prime }\subseteq \bigcup _{n=1}^{\infty }{E}_n\hfill \\ E_n\in B_0\hfill \end{array}\right\}=\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{{\mu }^{\prime }\left(\bigcup _{n=1}^{\infty }{E}_n\right)|\begin{array}{c}{B}^{\prime }\subseteq \bigcup _{n=1}^{\infty }{E}_n\hfill \\ E_n\in B_0\hfill \end{array}\right\}.$$

  For an arbitrary $𝜖>0$ approximate ${\mu }^{\text{∗}}(B^{\prime })$ by ${\mu }^{\prime }\left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n\right)$ satisfying

  $${\mu }^{\prime }\left(\bigcup _{n=1}^{\infty }{E}_n\right)\le {\mu }^{\text{∗}}(B^{\prime })+𝜖.$$

  By monotonicity of ${\mu }^{\prime }$ we then have

  $${\mu }^{\prime }(B)\le {\mu }^{\text{∗}}(B^{\prime })+𝜖.$$

  Sending $𝜖$ to zero, we obtain the desired result. Notice that we have not violated the possibility of $\mu (E)=\infty$ in any of the steps above.

• $\forall <!--\; FUNCTION\; APPLICATION\; -->E\in B\cap B^{\prime }:\mu (E)\le {\mu }^{\prime }(E)$.
  Case I. ${\mu }^{\text{∗}}(E)<\infty$.
  Let ${(E_n)}_{n\in N}$ be an outer cover of $E$ such that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{\mu }_0(E_n)-{\mu }^{\text{∗}}(E)\le \frac{𝜖}{2}$. By additivity of the measures, we then have:

  $$\begin{array}{llll}\hfill {\mu }^{\prime }\left(E\right)& ={\mu }^{\prime }\left(\bigcup _{n=1}^{\infty }{E}_n\right)-{\mu }^{\prime }\left(\bigcup _{n=1}^{\infty }{E}_n\setminus E\right).& \hfill & \\ \multicolumn{4}{c}{\text{By the previous part of this proof, we know that }{\mu }^{\text{∗}}\text{ is greater than or equal to}{\mu }^{\prime }\text{.}}\\ \\ \hfill & \ge {\mu }^{\prime }\left(\bigcup _{n=1}^{\infty }{E}_n\right)-{\mu }^{\text{∗}}\left(\bigcup _{n=1}^{\infty }{E}_n\setminus E\right)& \hfill & \\ \multicolumn{4}{c}{\text{By the construction of }{(E_n)}_{n\in N}\text{ we have an upper bound on how this cover differs from }E\text{:}}\\ \\ \hfill & \ge {\mu }^{\prime }\left(\bigcup _{n=1}^{\infty }{E}_n\right)-\frac{𝜖}{2}& \hfill & \\ \hfill & =\sum _{n=1}^{\infty }{\mu }^{\prime }(E_n)-\frac{𝜖}{2}& \hfill & \\ \hfill & \ge \sum _{n=1}^N{\mu }^{\prime }(E_n)-\frac{𝜖}{2}& \hfill & \\ \hfill & =\sum _{n=1}^N{\mu }^{\text{∗}}(E_n)-\frac{𝜖}{2}& \hfill & \\ \multicolumn{4}{c}{\text{Here we spend another }𝜖∕2\text{ and easily find a }N\in N\text{ such that the series}\sum _{n\in N}{\mu }^{\text{∗}}(E_n)\text{ is approximated by the finite sum }\sum _{n\le N}{\mu }^{\text{∗}}(E_n)\text{:}}\\ \\ \hfill & \ge \sum _{n=1}^{\infty }{\mu }^{\text{∗}}(E_n)-\frac{𝜖}{2}-\frac{𝜖}{2}& \hfill & \\ \hfill & \ge {\mu }^{\text{∗}}\left(\bigcup _{n=1}^{\infty }{E}_n\right)-𝜖& \hfill & \end{array}$$

  Sending $𝜖\to 0$, we obtain the desired result.
  Case II. ${\mu }^{\text{∗}}(E)=\infty$.
  This is where the $\sigma$-additivity property of $B_0$ becomes crucial. Let ${(X_m)}_{m\in \mathbb{N}}$ be the required sequence of sets in $B_0$, i.e., $X={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{m=1}^{\infty }{X}_n$ and ${\mu }_0(X_n)<\infty$. By the previous part of the proof, we have:

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->M\in \mathbb{N}:{\mu }^{\prime }\left(E\cap \bigcup _{m=1}^M{X}_m\right)\ge \mu \left(E\cap \bigcup _{m=1}^M{X}_m\right).$$

  Since both $\mu$ and ${\mu }^{\prime }$ are measures, we can send $M$ to infinity and, using the upwards monotone convergence, conclude that both quantities become equal in infinity.

• (Alternative proof) $\forall <!--\; FUNCTION\; APPLICATION\; -->E\in B\cap B^{\prime }:{\mu }^{\prime }(E)\le \mu (E)$.
  Alternatively, we can argue by contradiction: suppose that ${\mu }^{\prime }(E)>\mu (E)$. Then we can find ${(E_n)}_{n\in N}$ such that

  $${\mu }^{\prime }(E)>\sum _{n=1}^{\infty }{\mu }_0(E_n)$$

  contradicting the monotonicity of $E\subseteq {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n$.