relation between $C_0(X)$ and $L^{\infty}(X)$ Let $X$ be a locally compact topological space. If we want to consider $L^{\infty}(X)$ what measure I should take? Since it is a topological space it has a natural Borel $\sigma$ algebra generated by its open sets. How about the measure?
If $C_0(X)$ denotes the continuous functions vanishing at $\infty$ Can we have the inclusion $C_0(X)\subset L^{\infty}(X)$? 
Is $C_0(X)$ itself complete?
 A: $C_0(X)$ is a Banach space, namely the closure of $C_c(X)$, under the sup-norm. (Rudin: Real and Complex Analysis, 3.17) I am assuming $X$ locally compact Hausdorff (though the result is true for every topological space $X$ if you, instead, define $C_0(X)$ as the closure of $C_c(X)$).
It follows that $[C_0(X)]$ is a closed subspace of $L^\infty(\mu)$ for any positive Borel measure $\mu$ on $X$. (Because continuous functions are Borel-measurable and $C_0(X)$ functions are bounded, whichever definition you use.)
Here $[C_0(X)]:=\{[f]:\, f\in C_0(X)\}$, where $[f]$ is the equivalence class of $C_0$-functions equal to $f$ a.e.
The question remains whether $C_0(X)=[C_0(X)]$, or equivalently, is the mapping $f\mapsto [f]$, i.e., $C_0(X)\to L^\infty(X)$, injective, i.e., if $C_0(X)\owns f=0$ a.e., do we always have $f=0$.
This depends on your $\mu$. If $X\subset\mathbb R^n$ is, e.g., open (or an interval or a Cartesian product of intervals), then the Lebesgue measure $m$ will do: if $C_0(X)\owns f=0$ a.e.$[m]$, then $f=0$, by Lemma A below. Thus, then $C_0(X)\subset L^\infty(m)$ is a closed subspace.
Did this answer your question, or would you prefer more general sufficient conditions on $X$ for such a $\mu$ to exist?
Lemma A. If $f\in C(X)$, $f=0$ a.e.$[m]$, then $f=0$, if $X\subset \mathbb R^n$ is open or the Cartesian product of intervals of positive measure.
Proof: Let $f(x)\ne0$. By continuity, $f(x)\ne0$ on an open $x$-centric ball $B$ (say, on $B\cap X$ if $x$ is on the boundary of $X$). Then $m(B)>0$, a contradiction with $f=0$  a.e.[m], QED.
Intervals may be open, closed, or semi-open.
Example. Let $\mu(E)=1$ iff $0\in E$, else $\mu(E)=0$. Then $[f]=[0]$ iff $f(0)=0$, as then $\mu(\{x:f(x)\ne0\})\le \mu(\{0\}^c)=0$, i.e., $f=0$ a.e.$[\mu]$. So even on, say, $X=[0,1]$ not every measure $\mu$ satisfies our criterion. But $\mu=m$ will do. However, Lebesgue measure is defined only on measurable subsets of $\mathbb R^n$ (and on $\sigma$-algebras isomorphic to them).
