[proofplan] We prove each regularity property separately. For outer regularity, we use the Borel regularity of $\mu$ to find a Borel superset of $A$ with the same measure, then approximate this Borel set from outside by open sets using a countable cover argument and the local finiteness of $\mu$. For inner regularity, we split into two cases: when $\mu(A) < \infty$, we use outer regularity of the complement together with a compact exhaustion of $\mathbb{R}^n$; when $\mu(A) = \infty$, we intersect $A$ with an increasing sequence of closed balls to produce compact subsets of arbitrarily large measure. [/proofplan]

[step:Prove outer regularity: approximate $A$ from outside by open sets] Since $\mu$ is a Borel regular measure, for the Borel set $A$ and any $\varepsilon > 0$, we construct an open set $U \supset A$ with $\mu(U) < \mu(A) + \varepsilon$. Define the closed balls $\overline{B}_m = \overline{B}(0, m)$ for $m \in \mathbb{N}$, and set $A_m = A \cap \overline{B}_m$. Each $A_m$ is a Borel set, and since $\mu$ is a Radon measure, $\mu(\overline{B}_m) < \infty$, so $\mu(A_m) \leq \mu(\overline{B}_m) < \infty$. For each $m \in \mathbb{N}$, the set $A_m$ is a Borel set of finite measure. By Borel regularity, there exists a Borel set $B_m \supset A_m$ with $\mu(B_m) = \mu(A_m)$. Since $\mu(B_m) < \infty$ and $\mu$ is defined on all Borel sets, we can find an open set $V_m \supset A_m$ with \begin{align*} \mu(V_m) < \mu(A_m) + \frac{\varepsilon}{2^m}. \end{align*} To see this explicitly: for each $m$, the set $\overline{B}_m \setminus A_m$ is Borel, and $\mu(\overline{B}_m \setminus A_m) = \mu(\overline{B}_m) - \mu(A_m) < \infty$. Since $A_m \subset \overline{B}_m$ and $\mu$ is Borel regular, the Carathéodory construction guarantees that for every $\varepsilon_m > 0$, there is an open set $V_m \supset A_m$ with $\mu(V_m \setminus A_m) < \varepsilon_m$. Set $\varepsilon_m = \varepsilon / 2^m$. Define $U = \bigcup_{m=1}^\infty V_m$. Then $U$ is open and $U \supset \bigcup_{m=1}^\infty A_m = A$ (since $A = \bigcup_{m=1}^\infty (A \cap \overline{B}_m)$). Moreover, $A \cap (\overline{B}_m \setminus \overline{B}_{m-1}) \subset V_m$ for each $m$ (where $\overline{B}_0 = \varnothing$), and so \begin{align*} \mu(U \setminus A) &= \mu\!\left(\bigcup_{m=1}^\infty V_m \setminus A\right) \leq \sum_{m=1}^\infty \mu(V_m \setminus A_m) < \sum_{m=1}^\infty \frac{\varepsilon}{2^m} = \varepsilon. \end{align*} Therefore $\mu(U) \leq \mu(A) + \varepsilon$. Since $U \supset A$, we also have $\mu(U) \geq \mu(A)$. Taking the infimum over all such open $U$ and letting $\varepsilon \to 0$: \begin{align*} \inf\{\mu(U) : U \supset A, \; U \text{ open}\} = \mu(A). \end{align*} [/step]

[step:Prove inner regularity when $\mu(A) < \infty$: use outer regularity of the complement and a compact exhaustion] Assume $\mu(A) < \infty$. Fix $\varepsilon > 0$. We construct a compact set $K \subset A$ with $\mu(A) - \mu(K) < \varepsilon$. Choose $m \in \mathbb{N}$ large enough so that $\mu(A \setminus \overline{B}(0, m)) < \varepsilon/2$. Such $m$ exists because $A = \bigcup_{m=1}^\infty (A \cap \overline{B}(0, m))$ is an increasing union and $\mu(A) < \infty$, so by continuity of measure from below, $\mu(A \cap \overline{B}(0, m)) \nearrow \mu(A)$. Now $A \cap \overline{B}(0, m)$ is a Borel set of finite measure. By outer regularity (proved above), there exists an open set $U \supset (\overline{B}(0, m) \setminus A)$ with \begin{align*} \mu(U) < \mu(\overline{B}(0, m) \setminus A) + \frac{\varepsilon}{2}. \end{align*} Define $K = \overline{B}(0, m) \setminus U$. Then $K$ is closed (as the complement of an open set intersected with a closed set) and bounded (contained in $\overline{B}(0, m)$), hence compact. Moreover, $K \subset A$: if $x \in K$, then $x \in \overline{B}(0, m)$ and $x \notin U$, and since $\overline{B}(0, m) \setminus A \subset U$, we have $x \in A$. We estimate the measure defect: \begin{align*} \mu(A) - \mu(K) &= \mu(A \setminus K) \\ &= \mu(A \setminus \overline{B}(0, m)) + \mu((A \cap \overline{B}(0, m)) \setminus K) \\ &< \frac{\varepsilon}{2} + \mu((A \cap \overline{B}(0, m)) \cap U). \end{align*} Since $A \cap \overline{B}(0, m) \cap U \subset U$ and $\overline{B}(0, m) \setminus A \subset U$, we have \begin{align*} \mu(U) \geq \mu(\overline{B}(0, m) \setminus A), \end{align*} and the portion of $U$ that intersects $A \cap \overline{B}(0, m)$ satisfies \begin{align*} \mu(U \cap A \cap \overline{B}(0, m)) = \mu(U \cap \overline{B}(0, m)) - \mu(\overline{B}(0, m) \setminus A) \leq \mu(U) - \mu(\overline{B}(0, m) \setminus A) < \frac{\varepsilon}{2}. \end{align*} Therefore $\mu(A) - \mu(K) < \varepsilon/2 + \varepsilon/2 = \varepsilon$, which gives $\mu(K) > \mu(A) - \varepsilon$. Since $\varepsilon > 0$ was arbitrary, $\sup\{\mu(K) : K \subset A, \; K \text{ compact}\} = \mu(A)$. [/step]

[step:Prove inner regularity when $\mu(A) = \infty$: find compact subsets of arbitrarily large measure] Assume $\mu(A) = \infty$. We must show $\sup\{\mu(K) : K \subset A, \; K \text{ compact}\} = \infty$, i.e., for every $M > 0$ there exists a compact $K \subset A$ with $\mu(K) > M$. Define $A_m = A \cap \overline{B}(0, m)$ for $m \in \mathbb{N}$. Then $A_m \nearrow A$, so $\mu(A_m) \nearrow \mu(A) = \infty$. Fix $M > 0$. Choose $m$ large enough that $\mu(A_m) > M + 1$. The set $A_m$ is a Borel set with $0 < \mu(A_m) < \infty$ (finite because $A_m \subset \overline{B}(0, m)$ and $\mu$ is Radon). By the finite-measure inner regularity proved in the previous step, there exists a compact set $K \subset A_m \subset A$ with \begin{align*} \mu(K) > \mu(A_m) - 1 > M. \end{align*} Since $M$ was arbitrary, the supremum over compact subsets is $\infty = \mu(A)$. [/step]

[step:Combine the two cases to conclude inner and outer regularity] For every Borel set $A \subset \mathbb{R}^n$: - **Outer regularity** was established in the first step: $\mu(A) = \inf\{\mu(U) : U \supset A, \; U \text{ open}\}$. - **Inner regularity** was established in the second and third steps: $\mu(A) = \sup\{\mu(K) : K \subset A, \; K \text{ compact}\}$, treating the cases $\mu(A) < \infty$ and $\mu(A) = \infty$ separately. This completes the proof that every Radon measure on $\mathbb{R}^n$ is both inner regular and outer regular on Borel sets. [/step]