I'm studing the following theorem
The dual space of $(C[0,1], ||\cdot||_\infty)$ is isomorphic to the space $V[0,1]$ of bounded variations functions $g$ that are left continuos and $g(0)=0$ with the norm $||g||= Var_0^1 g$.
In particular the author considers a functional $F \in C[0,1]^*$, and he wants to show that for any $f\in C[0,1]$, $F(f) = \int_0^1 f(x) dg(x)$ for some $g\in V[0,1]$. So he considers $\tilde{F} \in L^{\infty}[0,1]^*$ that extends $F$ ($\tilde{F}$ exists thanks to the Hahn-Banach theorem) and he puts $g(x) = \tilde{F}(\chi_{[0,x)})$ ($\chi_I$ is the characteristic function of $I$). It turn out that $g$ is of bounded variations and $g(0)= 0$ but maybe it is not left continuous. In order to solve this issue, the author considers $\tilde{g}(x)= \lim_{\epsilon\to 0^+} g(x-\epsilon)$. The author claims that $\tilde{g}$ is left continuous and that the integral $\int_0^1 f(x) dg(x) = \int_0^1 f(x) d\tilde{g}(x)$.
1)What ensures us that the limit exists i.e. that $\tilde{g}$ is well defined?
2) Why $\int_0^1 f(x) dg(x) = \int_0^1 f(x) d\tilde{g}(x)$?
A real function $g$ of bounded variation on $[0,1]$ can be decomposed into two monotone functions. If $V_{0}^{x}(g)$ is the variation of $f$ on $[0,x]$, then $$ |g(x)-g(0)| \le V_{0}^{x}(g). $$ This gives a decomposition of $f(x)-f(0)$ into the difference of monotone non-decreasing functions: $$ g(x)- g(0) = V_{0}^{x}(g) - \{V_{0}^{x}(g)-(g(x)-g(0))\} $$ Monotone functions always have left- and right-hand limits that are found using inf and sup. So $g$ has left- and right-hand limits.
If $f$ is a continuous function on $[0,1]$ and $g$ is a monotone, then $$ \int_{0}^{1}fdg = \int_{0}^{1}fd\tilde{g} $$ where $\tilde{g}$ is any other monotone function for which $$ g(x^-) \le \tilde{g}(x) \le g(x^+). $$ And that's because $g(x^-)=\tilde{g}(x^-)$, $g(x^+)=\tilde{g}(x^+)$ for all points in $(0,1]$ and $[0,1)$, respectivey.