Let $\nu$ be a finite signed Borel measure on the closed interval $[a,b]$. We can define a function $F_\nu : [a,b]$ by $$F_\nu(x) = \nu([a,x]).$$It can be shown that $F_\nu$ has bounded variation and is right-continuous. An exercise in my class notes asks to prove that $|\nu|([a,b]) = V(F_\nu, [a,b])$, where the right-hand side denotes the total variation of $F_\nu$. Immediately after this, the notes state the following theorem:
"The map $\nu \mapsto F_\nu$ is a bijection from the set of finite signed Borel measures on $[a,b]$ to the set of functions of bounded variation that take value $0$ at $a$ and that are right-continuous."
My problem is that neither of these statements are true. Both break when we consider the measure $\delta$ given by $$\delta(A) = \begin{cases} 1, &\quad a \in A \\ 0, &\quad a \notin A \end{cases}.$$
In this case, $F_\delta$ is identically $1$, and so $V(F_\delta,[a,b]) = 0$ while $|\delta|([a,b]) = \delta([a,b]) = 1$. Moreover, $F_\delta$ does not take the value $0$ at $a$, so the second statement is false too.
How do we fix this? Can we just limit ourselves to the meausures $\nu$ for which $\nu(\{a\}) = 0$? Or is there a change we can make so that the results still apply to all signed measures?
