Why can the elements of $L^\infty$ be approximated in $L^p$ by $C^1$-functions?

Question

Let $\Omega\subseteq\mathbb R^n$ be a bounded domain and $f\in L^\infty(\Omega)$. From which theorem does the existence of $(f_k)_{k\in\mathbb N}\subseteq C^1(\overline\Omega)$ with $$f_k\stackrel{L^p(\Omega)}{\to}f$$ for some $p>n$ folow?

Answer 1

This is true for any $p\in[1,\infty)$.

First, since $\Omega$ is bounded, $L^\infty(\Omega)\subset L^p(\Omega)$. Say $\epsilon>0$. Choose $K\subset\Omega$ with $K$ compact, so that if $g=f\chi_K$ then $$||f-g||_p<\epsilon.$$

Now if $\phi_n$ is a smooth approximate identity with compact support then the convolution $g*\phi_n$ is smooth, $$||g-g*\phi_n||_p\to0,$$and if $n$ is large enough then $g*\phi_n$ is supported in a compact subset of $\Omega$.

Answer 2

you just need to show that characteristics of intervals can be approximated in $L^p$ norm, and then extend this result arbitrary measurable sets (using the regularity of the Lebesgue measure). Finally extend this result to simple functions.

This is also Lusin Theorem. Whose main idea is to use Urysohn's functions. So the answer of your why question is because of the regularity of the lebesgue measure which read as follows $$\forall \; C \subset \mathcal{B}(\Bbb{R}^n)\text{ and } \epsilon>0 \; \text{ there are $K_{\epsilon}$ compact and $A_\epsilon$ open such that } K_\epsilon\subset C \subset A_\epsilon\; \lambda(A_\epsilon \setminus K_\epsilon) <\epsilon$$