In general, the total variation of a measure (which appears in the identification of $C^*$, endowed with the operator norm, with the space of Radon measures, endowed with the total variation), defined for $\mu\in C_0(\Omega)^*$ as
$$|\mu|(\Omega) := \sup\left\{\sum_{i=0}^\infty |\mu(X_i)|:\bigcup_{i=0}^\infty X_i = \Omega\right\},$$
is not the same as the total variation of a function (which appears in the definition of the normed space $BV$), defined for $f\in BV(\Omega)$ as
$$ TV(f) := \sup \left\{\int_\Omega f (-\mathrm{div} \varphi) \,dx: \varphi\in(C_0^\infty(\Omega))^n,\, \sup_{x\in\Omega}|\varphi(x)|\leq 1\right\}.$$
You could say, however, that if a function has bounded variation, its distributional gradient exists as a Radon measure. The total variation (in the sense of the seminorm on $BV$) of the function is then the same as the total variation (in the sense of measures) of its distributional gradient.
However, in the one-dimensional case, the Riesz representation theorem actually does yield a function of bounded variation (this is in fact Riesz' original statement of 1909). In this case, the integration with respect to a function (of bounded variation) is in the sense of a Riemann-Stieltjes integral.
Best Answer
Let $f\in BV[a,b]$ Let $f=\phi_1 - \phi_2$ be the Jordan decomposition. We can now define two pre-measures $\mu_0'$ and $\mu_0''$. From which we construct the outer measures $\mu'^\star$ and $\mu''^\star$ and by Carathéodory's extension theorem we get $\mu'$ and $\mu''$. Now define $\mu=\mu'+\mu''$ by Jordan’s theorem for compositions of measures.
Our construction is obvoiusly injective. We will show that it is surjective as well. Let $\mu$ be a finite Borel sign-changing measure on $[a,b]$. Let $\mu=\mu^+-\mu^+$ be the Jordan decomposition. Set $\phi_1(x)=\mu^+([a,x))$ and $\phi_2(x)=\mu^-([a,x))$. Now $f=\phi_1 - \phi_2$ is a function of bounded variation as the difference of two monotonic non-negative such function.
We just established a bijection between the two sets. q.e.d.