> [!definition] > > Let $F: \real \to \complex$ be a function of [[Bounded Variation|bounded variation]]. $F \in NBV$ if the limit $F(-\infty) = 0$. > [!theorem] > > Let $\mu$ be a complex [[Borel Measure|Borel measure]] and $F(x) = \mu((-\infty, x])$, then $F \in NBV$. If $F \in NBV$, then the measure $\mu_F$ defined by $\mu_F((-\infty, x]) = F(x)$ is a Borel measure. > [!theorem] > > Let $F \in NBV$, then $F$ is **absolutely continuous** if for every $\eps > 0$, there exists $\delta > 0$ such that for any disjoint collection of intervals $\seqf{(a_j, b_j)}$, > $ > \sum_{j = 1}^n m((a_j, b_j)) < \delta \Rightarrow \sum_{j = 1}^n \abs{F(b_j) - F(a_j)} < \eps > $ > [!theorem] > > Let $F \in NBV$, then $F' \in L^1$. Moreover, $\mu_F \perp m$ if and only if $F' = 0$ a.e., and $\mu_F \ll m$ if and only if $F(x) = \int_{(-\infty, x]}F'$. > > *Proof*. By the [[Lebesgue Differentiation Theorem]], > $ > \mu_F = F'dm + (\mu_F - F'dm) > $ > [!theorem] > > Let $-\infty < a < b < \infty$ and $F: [a, b] \to \complex$, then the following are equivalent: > 1. $F$ is absolutely continuous on $[a, b]$. > 2. There exists $f \in L^1$ such that $F(x) - F(a) = \int_{[a, x]}f$. > 3. $F$ is differentiable a.e. on $[a, b]$ with $F' \in L^1([a, b], m)$, and $F(x) - F(a) = \int_{[a, x]}F'$.