# Equivalent Measures via Hahn-Kolmogorov and $\sigma-$finiteness

Hello Mathematics community. I am currently struggling with the following problem from Terry Tao's Introduction to Measure Theory textbook. It deals with pre-measures and the Hahn-Kolmogorov extension.

Let $\mu_0: \mathcal{B}_0 \to [0, \infty]$ be a pre-measure, let $\mu: \mathcal{B} \to [0, \infty]$ be the Hahn-Kolmogorov extension of $\mu_0$, and let $\mu' : \mathcal{B'} \to [0, \infty]$ be another countably additive extension of $\mu_0$. Suppose also that $\mu_0$ is $\sigma-$finite, which means that one can express the whole space $X$ as the countable union of sets $E_1,E_2 \ldots \in \mathcal{B}_0$ for which $\mu_0(E_n) < \infty$ for all $n$. Show that $\mu$ and $\mu'$ agree on their common domain of definition. In other words, show that $\mu(E)=\mu'(E)$ for all $E\in \mathcal{B} \cap \mathcal{B'}.$

I know that a pre-measure on a Boolean algebra $\mathcal{B}_0$ is a finitely additive measure $\mu_0: \mathcal{B}_0 \to [0,\infty]$ with the property that $\mu_0(\cup_{n=1}^{\infty}E_n) = \sum_{n=1}^{\infty}\mu_0(E_n)$ whenever $E_1,E_2, \ldots \in \mathcal{B}_0$ are disjoint sets such that $\cup_{n=1}^{\infty} E_n$ is in $\mathcal{B}_0$.

By virtue of the Hahn-Kolmogorov Theorem, we can extend this finitely additive pre-measure $\mu_0$ to a countably additive measure $\mu$.

I do not know how to start solving this problem. The only thing that comes to mind is to first consider $\mu'(E) \leq \mu^*(E)$ for all $E \in \mathcal{B'}$ where $\mu^*$ is an outer measure. But what does this achieve? As always, any help is greatly appreciated. Thanks in advance.

This problem comes out in section 1.7 of the text. The link to the free online version is provided below:

http://terrytao.files.wordpress.com/2011/01/measure-book1.pdf