Density in the uniform norm and in $L^p$ norm

Question

I read that the separability of $L^p(X)$ spaces with $1\leq p < +\infty$ follows from the separability of the space of continous functions on $X$ with compact support $C_o(X)$ which is dense in $L^p(X)$ in $L^p$ norm . Indeed there is a countable subset $D$ of $C_o(X)$ which is dense in $C_o(X)$ with the uniform norm. I can't understand why the density of $D$ in $C_o(X)$ in the uniform norm implies the density of $D$ in $C_o(X)$ in $L^p$ norm (and so the density of $D$ in $L^p(X)$). $X$ is an open subset of $R^n$ with the usual topology and Lebesgue measure.

Answer 1

I start by the case where $X$ has a finite measure.

Let $f$ be a function in $\mathbb L^p$ and let $\varepsilon\gt 0$. We would like to find an element $g$ of $D$ such that $\left\lVert f-g\right\rVert_p\leqslant\epsilon$. Using the density of $C_0$, we are able to find a function $g_1\in C_0$ such that $\left\lVert f-g_1\right\rVert_p\lt\epsilon/2$. Now, since $D$ is dense in $C_0(X)$ for the uniform norm, there exists a function $g\in D$ such that $$\sup_{x\in X}\left\lvert g_1(x)-g(x)\right\vert\leqslant \frac{\epsilon}{2\left(\lambda\left(X\right)\right)^{1/p}},$$ which implies that $\left\lVert g_1-g\right\rVert_p\leqslant\epsilon/2$ hence $\left\lVert f-g\right\rVert_p\leqslant\epsilon/2$.

This proves that $\mathbb L^p(X)$ is separable when $X$ is an open set of finite measure. To treat the general case, write $X$ as a countable union of open sets of finite measure (say $X=\bigcup_{k\geqslant 1}X_k$); find a countable dense subset $D_k$ of $\mathbb L^p\left(X_k\right)$ and check that $\bigcup_{k\geqslant 1}D_k$ is dense in $\mathbb L^p(X)$.