Regular Borel measures

Question

I'm trying to solve the following problem and got stuck.

Let $\mu$ and $\nu$ be two regular Borel measures on $\mathbb{R}^n$ such that $\mu = \nu$ on $\mathcal{B}(\mathbb{R}^n)$. Then $\mu = \nu$ on $\mathcal{P}(\mathbb{R}^n)$.

My idea was the following:

Let $F\subset\mathbb{R}^n$, then by definition there exists $E\in\mathcal{B}(\mathbb{R}^n)$ such that $F\subset E$ and

$$\mu(F) =\mu(E) = \nu(E) = \nu(H)$$

for some $H\subset\mathbb{R}^n$ and $H\subset E$

However, I don't know how to show that $F=H$. Any idea on how to show that $H = F$?

Thank you for the help.

Answer 1

For $F\subseteq\mathbb R^n$ sets $G,H\in\mathcal B(\mathbb R^n)$ exist with:

• $F\subseteq G$ and $\mu(F)=\mu(G)$
• $F\subseteq H$ and $\nu(F)=\nu(H)$

Then $K:=G\cap H\in\mathcal B(\mathbb R^n)$ with $F\subseteq K\subseteq G$ and $F\subseteq K\subseteq H$.

Based on that we find that $\mu(F)=\mu(K)$ and $\nu(F)=\nu(K)$ so that: $$\mu(F)=\mu(K)=\nu(K)=\nu(F)$$