The dual space of $C(Y,\mathbb R)$ when $Y$ is a complete and separable metric space Just to be confirmed what is the dual space of $C(Y,\mathbb R)$ i.e the vector space of all continuous functions $f: Y\to \mathbb R$ when $Y$ is complete and separable metric space? Is it the same when $Y$ is compact, which is the space of all signed/complex measures on $Y$? Thanks for any comment.
Even, I would be interested to know the dual space, when continuous functions are bounded or maybe even Lipschitz.

If you are also interested in the space $C_b(Y, \mathbb{R})$ of bounded continuous functions endowed with the topology of uniform convergence, which is normed by
\begin{equation*}
||f|| := \sup_{y \in Y} |f(y)|, \quad f \in C_b(Y, \mathbb{R}),
\end{equation*}
then the assumption that $Y$ is complete separable metric space is not sufficient for identifying the dual $C(Y,\mathbb{R})'$ with the family of finite Borel measures on $Y$. 
Here is a standard counterexample. Let $Y := \mathbb{N}$ and give it the discrete metric; on the other hand, let $C_0(\mathbb{N},\mathbb{R})$ be the subspace of $C_b(\mathbb{N},\mathbb{R})$ consisting of convergent sequence. Define a linear functional $L$ on $C_0(\mathbb{N}, \mathbb{R})$ as follows:
\begin{equation*}
L(f) := \lim_{n \to \infty} f(n), \quad f \in C_0(\mathbb{N},\mathbb{R}).
\end{equation*}
It is easy to see that $L$ is bounded and hence continuous. By virtue of the Hahn-Banach theorem, $L$ can be continuously extended over $C_b(\mathbb{N},\mathbb{R})$. However, $L$ cannot be identified with any countably additive measure on $\mathbb{N}$.
