Lemma 17.1.6. Let $X$ be a LCH space, $Y$ be a compact Hausdorff space, $\mu$ be a finite Radon measure on $X \times Y$, and $\nu$ be a measure on $X$. If for each $A \in \cb_{X}$, $\nu(A) \le \mu(A \times Y)$, then $\nu$ is also a Radon measure.
Proof. Let $A \in \cb_{X}$ and $\eps > 0$. By outer regularity of $\mu$, there exists $U \in \cn_{X \times Y}(A \times Y)$ such that $\mu(U \setminus (A \times Y)) < \eps$. By the Tube Lemma, there exists $V \in \cn_{X}(A)$ such that $V \times Y \subset U$. In which case,
so $\nu$ is outer regular on $A$.
On the other hand, by Proposition 17.1.3, there exists $K \subset A \times Y$ compact such that $\mu((A \times Y) \setminus K) < \eps$. By Proposition 4.16.2, $\pi_{1}(K) \subset A$ is also compact. Thus
so $\nu$ is inner regular on $A$.$\square$