Lusin's theorem says that in a finite measure space, given a measurable function $\varphi$, for every $\varepsilon \gt 0$ there exists a continuous function $g$ such that $$ \mu\left(\{x : \varphi(x)\neq g(x)\}\right)\lt \varepsilon.$$
How can I use this to show that every $f\in L^p$ can be approximated in $L^p$ by continuous functions of compact support?
Best Answer
I'm not sure you're stating all conditions of Lusin's theorem correctly - the one I'm looking at (from Real Analysis by Gerald Folland) assumes that $\mu$ is a Radon measure on a locally compact Hausdorff space $X$, and concludes that there exists a compactly supported function $g$ that coincides with the given $\varphi$ on a set of measure $< \varepsilon$. If $\varphi$ is bounded, then $g$ can be chosen so that $\| g \|_\infty \leq \| \varphi \|_\infty$.
Anyways, here are some hints: it suffices to show that any characteristic function $\chi_E$ (where $E \subseteq X$) can be approximated arbitrarily well in $L^p$-norm by compactly supported functions (since simple functions are dense in $L^p$ for $1 \leq p < \infty$). Showing this is a simple matter of estimating $$ \int_X |\chi_E(x) - \varphi(x)|^p \,\mathrm{d}\mu(x)$$ Note that, since $\chi_E$ is bounded, we can choose $g$ such that $\| g \|_\infty \leq \|\chi_E \|_\infty = 1$, and hence $\| \chi_E - \varphi \|_\infty \leq 2$.