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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?

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1 Answer 1

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 $\| \chi_E - \varphi \|_\infty \leq 2$.

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are you saying it is enough to show that simple functions are dense in $L^p$? –  Kuku May 4 '12 at 17:17
    
@kuku: Not entirely, no: you still need that small argument I've outlined for you, which essentially shows that $C_c(X)$ is dense in the simple functions. But if you don't already know that simple functions are dense in $L^p$ (in other words, if this fact has not been stated previously in your textbook/course), then there might be less complicated ways of doing it (not that the proof is particularily lengthy - Folland proves density of simple functions in $L^p$ in 5 lines or so). –  Martin Wanvik May 4 '12 at 17:29
    
@kuku: I've edited my answer. It was somewhat misleading, we obviously need an estimate that is independent of the chosen function $\varphi$, and so it is not sufficient to conclude that $\| \chi_E - \varphi\|_\infty $ is finite. –  Martin Wanvik May 4 '12 at 17:55

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