Prove a Borel measure coincides with a Riesz measure on Borel $\sigma$-algebra

Question

I try to prove the Lemma used here :Borel measure and Riesz measure

To prove:

If a Borel measure $\mu$ coincides with a Riesz measure $\lambda$ on any open set in $\mathbb{R^n}$, then they coincides on Borel $\sigma$-algebra $\mathcal{B}$.

My attempt:

Let $A\in \mathcal{B}$, for $\epsilon >0$, there exists $F\subset A\subset E$, with $F$ closed and $E$ open such that $\mu(E\setminus F)<\epsilon$. Since $E$ and $E\setminus F$ are open, we have:

$\mu(A)=\mu(E)-\mu(E\setminus A)\geq\mu(E)-\mu(E\setminus F)=\lambda(E)-\lambda(E\setminus F)\geq \lambda(A)-\epsilon$.

We conclude $\lambda(A)=\mu(A)$ by reserving $\mu$ and $\lambda$ and let $\epsilon \rightarrow 0 $.

My confusion is that

1. the existence of such $E$ and $F$ requires $\mu$ or $\lambda$ to be $\sigma$-finite. In this case, a Borel measure and a Riesz measure may not be $\sigma$-finite.
2. $\mu(E)-\mu(E\setminus A)$ may be undefined.

Any help would be appreciated.

Edit: The existence of $E$ and $F$ are clear to me now. Since Riesz measure is finite on any compact set, and $\mathbb{R^n}$ is a countable union of an ascending sequence of compact sets, we have $\sigma$-finiteness.

Answer 1

One way to approach this is via a generating class argument:

Define $\mathcal{A}=\{S\in\mathcal{P}(\mathbb{R}^{n}) \space|\space \lambda(S) = \mu(S)\}$. The objective is to show that $\mathcal{A}\supseteq\mathcal{B}$. By what's given, we already know that $\mathcal{A}$ contains the open sets in $\mathbb{R}^{n}$. Thus, since $\mathcal{B}$ is the smallest $\sigma$-algebra containing the open sets, the result will follow once we establish $\mathcal{A}$ is a $\sigma$-algebra. Since $\lambda$ and $\mu$ are both measures, it is immediate that $\emptyset \in \mathcal{A}$. It thus remains to show $\mathcal{A}$ is closed under complements and countable unions. This last part is where the work lies.