Proof [Theorem 7.10, Fol99]. First assume that $f$ is bounded.

(1, bounded): If $f$ is bounded, then $f \in L^{1}(X; E)$. By Proposition 17.1.7, there exists $\seq{\phi_n}\subset C_{c}(X)$ such that $\phi_{n} \to f$ in $L^{1}(\mu)$. Since $\phi_{n} \to f$ in $L^{1}(\mu)$, $\phi_{n} \to f$ in measure by Proposition 12.1.12. By taking a subsequence using 18.6.4, assume without loss of generality that $\phi_{n} \to f$ almost everywhere.

By Egoroff’s Theorem, there exists $A \subset \bracs{f \ne 0}$ such that $\phi_{n} \to f$ uniformly and $\mu(\bracs{f \ne 0}\setminus A) < \eps/3$. By Proposition 17.1.3, there exists $K \subset A$ compact such that $\mu(A \setminus K) < \eps/3$.

Let $\phi_{0} = \lim_{n \to \infty}\phi_{n}$ on $K$, then $\phi_{0} \in C(K; E)$ by Proposition 6.2.2, $\phi_{0} = f$ almost everywhere on $K$, and $\mu(\bracs{f \ne 0}\setminus K) < \eps$.

(2, bounded): By outer regularity, there exists $U \in \cn^{o}(\bracs{f \ne 0})$ such that $\mu(U \setminus \bracs{f \ne 0}) < \eps/3$.

By the Tietze Extension Theorem, there exists $\phi \in C_{c}(X; \complex)$ such that $\phi|_{K} = \phi_{0}$ and $\supp{\phi}\subset U$. In which case,

(3): Let

then $\psi$ is continuous with $\bracs{\phi = f}= \bracs{\psi \circ \phi = f}$ and $\norm{\psi \circ \phi}_{u} \le \norm{f}_{u}$.

Now assume that $f$ is arbitrary.

(1, unbounded): Since $\mu\bracs{f \ne 0}< \infty$, there exists $\alpha > 0$ such that $\mu\bracs{|f| > \alpha}< \eps/2$. Let $g \in L^{\infty}(X; E)$ such that $g = f$ on $\bracs{|f| \le \alpha}$. By (1) applied to $g$, there exists $A \subset \bracs{|f| \le \alpha}$ such that $f|_{A}$ is continuous and $\mu(\bracs{|f| \le \alpha}\setminus A) < \eps/2$. In which case, $\mu(\bracs{f \ne 0}\setminus A) < \eps$, as desired.

(2, unbounded): By (1) applied to $g$, there exists $\phi \in C_{c}(X; \complex)$ such that $\mu\bracs{\phi \ne g}< \eps/2$, and $\mu(\bracs{\phi \ne f}) < \eps$.$\square$