> [!theorem] > > Let $F: \real \to \real$ be a non-decreasing [[Function|function]] and $G(x) = \lim_{y \downto x}F(x)$. > 1. $F$ is [[Continuity|continuous]] on all but countably many points. > 2. $F$ and $G$ are [[Derivative|differentiable]] [[Almost Everywhere|a.e.]], with $F' = G'$ a.e. > > *Proof*. Since $G$ is non-decreasing and right-continuous, $G$ induces a regular [[Lebesgue-Stieltjes Measure]] $\mu_G$ on $\real$, where > $ > G(x + h) - G(x) = \begin{cases} > \mu_G((x, x + h]) & h > 0 \\ > -\mu_G((x + h, x]) & h < 0 > \end{cases} > $ > Since $(x, x + h]$ and $(x + h, x]$ both shrink nicely to $x$ as $h \to 0$, by the [[Lebesgue Differentiation Theorem]], $G$ is differentiable a.e. > > Let $H = G - F$, and $\seq{x_j}$ be an enumeration of points at which $H \ne 0$, then $H(x_j) > 0$ for all $j$. Let $j_H = \sum_{j \in \nat}H(x_j)\delta_{x_j}$, then $j_H$ is $\sigma$-finite and finite on compact sets, thus regular. Moreover, $j_H$ By the Lebesgue Differentiation Theorem again, > $ > \abs{\frac{H(x + h) - H(x)}{h}} \le 4\abs{\frac{j_H((x - 2\abs{h}, x + 2\abs{h}))}{4\abs{h}}} \to 0 > $ > as $h \to 0$ for a.e. $x$. Thus $H' = 0$ a.e. > [!definition] > > Let $F: \real \to \complex$, and define > $ > T_F(x) = \sup\bracs{\sum_{j = 1}^n\abs{F(x_j) - F(x_{j - 1})}: n \in \nat, -\infty < x_j < x_{j + 1} \le x} > $ > as the **total variation function** of $F$, which is a non-negative, non-decreasing function in $x$. Moreover, > $ > \begin{align*} > \norm{F}_{v, [a, b]} &= \sup\bracs{\sum_{j = 1}^n\abs{F(x_j) - F(x_{j-1})}: n \in \nat, a \le x_j \le x_{j + 1} \le b} \\ > &= T_F(b) - T_F(a) > \end{align*} > $ > is the **total variation** of $F$ on $[a, b]$, and can be computed in terms of the total variation function if $T_F(a) < \infty$. > > If $\norm{F}_{v, [a, b]} < \infty$, then $F \in BV([a, b])$ is of **bounded variation**, and $\norm{ \cdot }_{v, [a, b]}$ is a [[Seminorm|seminorm]] on $BV([a, b])$. If $F \in BV([a, b])$ for every bounded interval, then $f \in BV_{\text{loc}}(\real)$. > [!theorem] > > Let $F: \real \to \real$ with $F \in BV$, then $T_F + F$ and $T_F - F$ are non-decreasing. > > *Proof*. Let $-\infty < x < y < \infty$ and $\eps > 0$, then there exists a partition $-\infty < x_0 < \cdots < x_n = x$ such that > $ > \sum_{j=1}^n \abs{F(x_j) - F(x_{j-1})} \ge T_F(x) - \eps > $ > From here, > $ > \begin{align*} > F(y) &= F(y) - F(x) + F(x) \\ > T_F(y) \pm F(y) &\ge \sum_{j = 1}^n \abs{F(x_j) - F(x_{j - 1})} + \abs{F(y) - F(x)} \\ > &\pm[F(y) - F(x)] \pm F(x) \\ > &\ge T_F(x) \pm F(x) - \eps > \end{align*} > $ > [!theorem] > > Let $F: \real \to \real$. If $F$ is monotone, then $F \in BV$ if and only if it is bounded. If $F$ is arbitrary, then $F \in BV$ if and only if there exists bounded non-decreasing functions $F^+, F^- \in BV$ such that $F = F^+ - F^-$. If $F \in BV$, then they can be taken to be > $ > F^+ = \frac{1}{2}(T_F + F) \quad F^- = \frac{1}{2}(T_F - F) > $ > which is known as the **Jordan decomposition** of $F$. > [!theorem] > > Let $F: \real \to \complex$. If $F \in BV$, then > 1. The limits $F(x-)$, $F(x+)$, $F(-\infty)$, $F(\infty)$ exists for all $x \in \real$. > 2. The set of points at which $F$ is discontinuous is countable. > 3. If $G(x) = F(x+)$, then $F'$ and $G'$ exists and are equal almost everywhere. > > *Proof*. Follows from the Jordan decomposition. > [!theorem] > > Let $F \in BV$, then $T_F(-\infty) = 0$. If $F$ is right-continuous, then so is $T_F$. > > *Proof*. Suppose that $F$ is right-continuous. Let $\Delta = T_F(x+) - T_F(x)$ and $\eps > 0$, then there exists $\delta > 0$ such that $\abs{F(x + h) - F(x)} < \eps$ and $T_F(x + h) - T_F(x+) < \eps/2$ for all $h \in (0, \delta)$. > > Let $x < x_0 < \cdots < x_n = x + h$ such that > $ > \sum_{j = 1}^n \abs{F(x_j) - F(x_{j - 1})} \ge \frac{1}{2}[T_F(x + h) - T_F(x)] \ge \Delta > $ > so > $ > \sum_{j = 2}^n \abs{F(x_j) - F(x_{j - 1})} \ge \frac{1}{2}\Delta - \abs{F(x_1) - F(x_0)} \ge \Delta - \eps > $ > Let $x = y_0 < \cdots < m_m = x_1$ such that > $ > \sum_{j = 1}^m \abs{F(y_j) - F(y_{j - 1})} \ge \frac{1}{2}\Delta > $ > then > $ > \begin{align*} > \Delta + \eps &> T_F(x + h) - T_F(x) \\ > &\ge \sum_{j = 2}^n \abs{F(x_j) - F(x_{j - 1})} + \sum_{j = 1}^m \abs{F(y_j) - F(y_{j - 1})} \\ > &\ge 2\Delta -\eps \\ > \eps &> \Delta > \end{align*} > $ > so $\Delta = 0$.