When $L^p$ is separable? I have never studied any measure theory, so apologise in advance for my question. Let $\Omega$ be an open and bounded subset of $\mathbb{R}^d$. Denote $\Gamma=\partial\Omega$ and $\Gamma=\Gamma_1\cup\Gamma_2$. Let $H=L^2(\Omega,\mathbb{R}^d)$. Consider
$$V=\{v\in H\,|\, v=0\, \mbox{a.e. on}\, \Gamma_1\}.$$
The equality $v=0$ is understood in the sense of trace operator. Is $V$ separable? Is there any general way to determine separability of $L^p$ spaces?
 A: First of all, there is no non-zero continuous trace operator on $L^{2}(\Omega;\mathbb{R}^{d})$.  Unfortunately, this means $H$ isn't well-defined.  
If $\Omega$ is an open set, then $L^{p}(\Omega;\mathbb{R}^{d})$ (with respect to Lebesgue measure) is separable.  First, consider the case $d = 1$.  One way to see this is to write $\Omega = \bigcup_{n = 1}^{\infty} G_{n}$, where $\{G_{n}\}_{n \in \mathbb{N}}$ is an increasing sequence of compact subsets of $\Omega$, and observe that $C(G_{n})$ continuously embeds in $L^{p}(G_{n})$.  Explicitly, if $f \in C(G_{n})$, then
$$\|f\|_{L^{p}(\Omega)} \leq |\Omega|^{\frac{1}{p}} \|f\|_{C(G_{n})}.$$  It's a general fact of real analysis that $C_{c}(\Omega)$ is dense in $L^{p}(\Omega)$.  Thus, if $\{\psi_{m,n}\}_{n \in \mathbb{N}}$ is a dense subset of $C(G_{n})$, then the fact that any $\varphi \in C_{c}(\Omega)$ can be approximated arbitrarily well with linear combinations of $\{\psi_{m,n}\}_{m \in \mathbb{N}}$ for high enough $n$ implies the same for functions in $L^{p}(\Omega)$.  In particular, $\{\psi_{m,n}\}_{m,n \in \mathbb{N}}$ is dense in $L^{p}(\Omega)$.  
If $d > 1$, then $L^{p}(\Omega; \mathbb{R}^{d}) \simeq L^{p}(\Omega)^{\oplus d}$, where the isomorphism is given explicitly by
$$(f_{1},f_{2},\dots,f_{d}) \mapsto f_{1} e_{1} + f_{2} e_{2} + \dots + f_{d} e_{d}.$$
Thus, the fact that $L^{p}(\Omega)$ is separable implies that $L^{p}(\Omega;\mathbb{R}^{d})$ is separable as well.
