I am currently working on the following problem from a past qualifying exam.
Let $\nu$ be a finite signed measure on $\mathbb{R}.$ Define a function $$g(x)=\int_{-\infty}^{x}\:d\nu:=\int \chi_{(-\infty, x]\:d\nu}$$ where $\chi$ is the characteristic function.
(1) Prove that $g$ is right-continuous. That is $$\lim_{h\searrow 0}g(x+h)=g(x)$$ for all $x\in \mathbb{R}.$
(2) Consider the Radon-Nikodym decomposition of $\nu=fm+\mu,$ where $m$ is the Lebsegue measure, $f\in L^1(\mathbb{R})$, and $v\perp m.$ Prove that $g$ is differentiable at almost all $x\in \mathbb{R}$ and $g'(x)=f(x),m\text{-a.e. }x$
I am completely stuck on this problem and don't know how to proceed. Any help will be appreciated. Thanks in advance.
For (1), one needs to use the upper-continuity of measures. Since $g(x) = \nu((-\infty, x])$ any sequence $h_n$ with $h_n \searrow x$ obeys $$(-\infty, x] = \bigcap_{n\geq 1} (-\infty, x + h_n],$$ and $(-\infty, h_{n+1}] \subset (-\infty, h_n]$ for all $n\geq 1$. Thus $|\nu| < \infty$ implies that $$\nu((-\infty, x]) = \nu\left(\bigcap_{n\geq 1}(-\infty, x + h_n]\right) = \lim_{n\to \infty} \nu((-\infty, h_n]),$$ and as $h_n$ was any sequence decreasing to $x$ part (1) follows.
For (2), note that $$\frac{g(x+h/2) - g(x-h/2)}{h} = \frac{\nu((x-h/2, x+h/2])}{h} = \frac{\nu((x-h/2, x+h/2])}{m((x-h/2, x+h/2])}.$$ The family of intervals $\{(x-h/2, x+h/2]\}$ shrinks nicely to $x$ as $h\searrow 0$, so we can use the well-known theorem (see, for example, Folland's Real Analysis 2nd ed, Theorem 3.22) that $$\lim_{h\to 0} \frac{\nu((x-h/2, x+h/2])}{m((x-h/2, x+h/2])} = f(x)$$ for $m$-almost every $x\in \mathbb R^n$. A similar argument shows the same thing if $h\nearrow 0$. But this is equivalent to asserting that $g'$ exists $m$-a.e., and that $g' = f$ $m$-a.e.