[proofplan] The argument combines two results: [Luzin's Theorem](/theorems/12) extracts a compact set $K\subset A$ of measure within $\varepsilon$ of $\mu(A)$ on which $f$ is already continuous, and the [Continuous Extension Theorem](/theorems/11) extends the restriction $f|_{K}$ to a continuous map $\bar{f}:\mathbb{R}^{n}\to\mathbb{R}^{m}$ defined on all of $\mathbb{R}^{n}$. Since $\bar{f}$ agrees with $f$ on $K$, the set where the two functions disagree inside $A$ is contained in $A\setminus K$, which has measure less than $\varepsilon$. [/proofplan] [step:Extract a compact set of near-full measure where $f$ is continuous via Luzin's Theorem] We verify the hypotheses of [Luzin's Theorem](/theorems/12): $\mu$ is a Borel regular measure on $\mathbb{R}^{n}$, $f:\mathbb{R}^{n}\to\mathbb{R}^{m}$ is $\mu$-measurable, and $A\subset\mathbb{R}^{n}$ is $\mu$-measurable with $\mu(A)<\infty$ — all given by assumption. Therefore [Luzin's Theorem](/theorems/12) supplies a compact set $K\subset A$ satisfying: (i) $\mu(A\setminus K)<\varepsilon$, (ii) the restriction $f|_{K}:K\to\mathbb{R}^{m}$ is continuous. [guided] The goal of this step is to locate a "well-behaved core" inside $A$ — a compact set on which $f$ is already continuous, so that the approximation problem reduces to extending a continuous function. [Luzin's Theorem](/theorems/12) is the precise tool for this. It requires three hypotheses: 1. **Borel regularity of the measure.** The theorem's proof relies on inner regularity to approximate measurable sets from within by compact sets. Our measure $\mu$ is Borel regular by assumption, so this condition is met. 2. **$\mu$-measurability of $f$.** The proof approximates $f$ by simple functions whose level sets must be $\mu$-measurable. Since $f$ is $\mu$-measurable by hypothesis, this is satisfied. 3. **Finite measure of $A$.** The $\varepsilon$-room argument in Luzin's proof subtracts measure from $\mu(A)$ at each approximation step and requires $\mu(A)<\infty$ to ensure the total error remains controlled. This is given. With all three conditions verified, [Luzin's Theorem](/theorems/12) produces a compact set $K\subset A$ such that $\mu(A\setminus K)<\varepsilon$ and $f|_{K}$ is continuous. [/guided] [/step] [step:Extend $f|_{K}$ continuously to all of $\mathbb{R}^{n}$] The set $K$ is compact and the map $f|_{K}:K\to\mathbb{R}^{m}$ is continuous. By the [Continuous Extension Theorem](/theorems/11), there exists a continuous map \begin{align*} \bar{f}:\mathbb{R}^{n}&\to\mathbb{R}^{m} \end{align*} satisfying $\bar{f}(x)=f(x)$ for every $x\in K$. [/step] [step:Verify the extension agrees with $f$ except on a set of measure less than $\varepsilon$] Since $\bar{f}|_{K}=f|_{K}$, every point $x\in A$ at which $\bar{f}(x)\neq f(x)$ must lie in $A\setminus K$. Therefore \begin{align*} \{x\in A\mid \bar{f}(x)\neq f(x)\}\subset A\setminus K, \end{align*} and monotonicity of $\mu$ gives \begin{align*} \mu\!\bigl(\{x\in A\mid \bar{f}(x)\neq f(x)\}\bigr)\le\mu(A\setminus K)<\varepsilon. \end{align*} This establishes the desired approximation. [/step]