> [!theorem] > > Let $X$ be a [[Locally Compact Hausdorff Space|LCH]] space, $\mu$ be a [[Radon Measure|Radon measure]] on $X$, and $f: X \to \complex$ be a [[Measurable Function|measurable function]] that vanishes outside a set of finite measure, and $\eps > 0$. Then there exists a [[Compactly Supported|compactly supported continuous]] function $\phi \in C_c$ such that $\phi = f$ except on a set of measure $\eps$ or less. > > If $f$ is bounded, $\phi$ can be taken such that $\norm{\phi}_u \le \norm{f}_u$. > > *Proof*. Let $E = \bracs{f \ne 0}$ and suppose that $f$ is bounded, then there exists $\seq{g_n} \subset C_c(X)$ such that $g_n \to f$ in $L^1$. By taking a subsequence, assume without loss of generality that $g_n \to f$ in $L^1$ and [[Almost Everywhere|a.e.]] By [[Egoroff's Theorem]], there exists $A \subset E$ such that $\mu(E \setminus A) < \eps/3$ and $g_n \to f$ uniformly on $A$. > > Let $K \subset A \subset U$ such that $\mu(U \setminus A), \mu(A \setminus K) < \eps/3$. Since $g_n \to f$ uniformly on $B$, $f|_B$ is continuous. By the [[Tietze Extension Theorem]], there exists $h \in C_c(X)$ such that $h|_K = f|_K$ with $\supp{h} \subset U$, and $\norm{h} \le \norm{f|_K}_u \le \norm{f}_u$. On the other hand, $\bracs{f \ne h} \subset U \setminus K$, which has measure $\eps$ or less. > > Now suppose that $f$ is unbounded. Let $\delta > 0$ such that $A = \bracs{\abs{f} < \delta}$ has measure $\eps/2$ or less. Let $\phi \in C_c(X)$ such that $\phi|_{A^c} = f|_{A^c}$ except on a set of measure $\eps$ or less, this gives the desired function.