Uniqueness property for the space of finite measure. Let $\mu$ be a finite measure on $\mathbb{R}$ satisfying,
$$\int_{\mathbb{R}}f(x)d\mu(x)=0,~\forall f\in C_c(\mathbb{R})$$
Then is it true that $\mu =0$?
We know that the result is true for $L^1(\mathbb{R})$, which is a subspace of the above.
Edit after the comments of Kavi Rama Murthy Sir:
Is the result also true for complex measure $\mu$ on $\mathbb{R}$?
 A: For all $n=1,2,3,\dots$ there exists a non-negative function $f \in \mathrm{C}_{\mathrm{c}}(\mathbb{R})$ such that $f=1$ on $[-n,n]$. Therefore, for all $n=1,2,3,\dots$
\begin{equation}
\mu([-n,n]) \leq \int_\mathbb{R} f d \mu = 0.
\end{equation}
By countable subadditivity $\mu(\mathbb{R})=0$, and we are done.
A: If we prove this for real measures the result for complex measures follows by just taking real and imaginary parts. Any real measure is the difference of two positive finite measures. So what we have to prove is the following:
if $\mu$ and $\nu$ are two finite positive measures such that $\int f d\mu=\int f d\nu$ for all $f \in C_c(\mathbb R)$ then  $\mu =\nu$.
For this note that given $\epsilon >0$ we can find $N$ such that $\mu ([-N,N])^{c}) <\epsilon$ and $\nu ([-N,N])^{c}) <\epsilon$. There exists $f \in C_c(\mathbb R)$ such that $0 \leq f(x) \leq 1$ for all $x$ and $f(x)=1$ for $|x| \leq N$. If $g$ is any bounded continuous function then $fg \in C_c(\mathbb R)$ so $\int fg d\mu=\int fg d\nu$. I now leave it to you put these together to conclude that $\int g d\mu=\int g d\nu$. The fact that this hods for all bounded continuous$f$ implies that $\mu =\nu$. [This is standard and it is part of Portmanteus's Theorem characterizing weak convergence of probability measures]. 
