Functions of bounded variation as the dual of $C([a,b])$ I am trying to understand this proposition about the dual of $C([a,b])$. I would like some help with the following:
(1) What does the integral with respect to a function of bounded variation mean? 
(2) According to Riesz Representation Theorem, the dual of $C([a,b])$ is the space of regular Borel measures (Radon measures) on $[a,b]$. So what is the relation between these measures and the bounded variation functions?
It would be nice to get some explanation in terms of distributions.
 A: 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.
