Is it true that $C_0^\ast[0,+\infty) = NBV_{loc}[0,+\infty)$ Any function from $BV_{\operatorname{loc}}[0,+\infty)$ defines a continuous linear functional on $C_0[0,+\infty)$. But is it true that any continuous linear functional on $C_0[0,+\infty)$ is given by a function with variation bounded on compacts? Is it possible to say then that $C_0^\ast[0,+\infty) = NBV_{\operatorname{loc}}[0,+\infty)$, where "N" stays for normalisation $u(0)=0$?
For classes $BV_{\operatorname{loc}}[0,1] = BV[0,1]$ and $C_0[0,1] = C[0,1]$ this statement is a special case of the Riesz representation theorem.
 A: Fix a homeomorphism of $[0,1)$ onto $[0,+\infty)$, such as $\phi(x)=x/(1-x)$. For any $f\in C_0[0,+\infty)$ the composition $f\circ \phi$ is continuous on $[0,1)$ and has limit $0$ at $x=1$. Thus, it extends to a function $g\in C[0,1]$ such that $g(1)=0$. 
It is easy to see that the map $f\mapsto f\circ \phi$ is an isometric isomorphism between $C_0[0,+\infty)$ and the subspace $M=\{g\in C[0,1]: g(1)=0\}$. The linear functionals on $M$ extend to $C[0,1]$. Applying the Riesz representation theorem and translating its statement back to $M$, you will find that $M^*$ is the set of finite measures on $[0,1]$ that give mass $0$ to $\{1\}$. These precisely correspond  to the $NBV$ functions on $[0,1]$ that are continuous at $1$. You can then transplant these functions to $[0,\infty)$.

Remark: $NBV$ means not only normalization, but also one-sided continuity at the interior poins of the interval (textbooks differ on whether to require continuity from the left or from the right). One-sided continuity prevents "atom-splitting" such as 
$$u(x)=\begin{cases}0, \quad &x<1/2\\ 1/3, &x=1/2 \\ 1& x>1/2 \end{cases}$$
The derivative of this function $u$ is the unit point mass at $1/2$. The value of $u(1/2)$ could be different, and we'd still get the same measure $u'$. One-sided continuity allows us to have one-to-one correspondence between finite signed measures and functions.
