Jordan Decomposition

First prove additivity of variation: for $a \le x_1 < x_2 \le b$, $V_a^{x_2}(f) = V_a^{x_1}(f) + V_{x_1}^{x_2}(f)$.

The inequality $\le$ goes as follows: given any partition $P$ of $[a, x_2]$, if $x_1$ is absent we may insert it, since the sum can only grow ($\lvert f(v) - f(u)\rvert \le \lvert f(x_1) - f(u)\rvert + \lvert f(v) - f(x_1)\rvert$).

A partition containing $x_1$ has its sum split into a sum over $[a, x_1]$ plus a sum over $[x_1, x_2]$, each bounded by the variation on that interval; taking the supremum over $P$ gives $\le$.

The reverse $\ge$: concatenating a partition $P_1$ of $[a, x_1]$ with a partition $P_2$ of $[x_1, x_2]$ gives a partition $P_1 \cup P_2$ of $[a, x_2]$ whose sum is the sum of the two sums.

Taking suprema on each interval gives $\ge$.

All suprema are finite since $f$ is BV.

Now $g(x) = V_a^x(f)$ is increasing: for $x_1 < x_2$, additivity gives $g(x_2) = g(x_1) + V_{x_1}^{x_2}(f) \ge g(x_1)$ (variation is nonnegative).

And $h(x_2) - h(x_1) = (g(x_2) - g(x_1)) - (f(x_2) - f(x_1)) = V_{x_1}^{x_2}(f) - (f(x_2) - f(x_1))$; but $V_{x_1}^{x_2}(f) \ge \lvert f(x_2) - f(x_1)\rvert \ge f(x_2) - f(x_1)$ from the single partition $\{x_1, x_2\}$, so $h(x_2) - h(x_1) \ge 0$ — $h$ is increasing too.

The identity $f = g - h$ is the definition of $h$.

Take any partition $a = x_0 < x_1 < \dots < x_n = b$ of $[a, b]$.

On each piece, $\lvert f(x_i) - f(x_{i-1})\rvert = \lvert (g(x_i) - g(x_{i-1})) - (h(x_i) - h(x_{i-1}))\rvert \le \lvert g(x_i) - g(x_{i-1})\rvert + \lvert h(x_i) - h(x_{i-1})\rvert$ (triangle inequality). Since $g$ and $h$ are increasing, each difference is nonnegative, so the absolute values drop away.

Summing over $i = 1, \dots, n$, $\sum_{i=1}^{n} \lvert f(x_i) - f(x_{i-1})\rvert \le \sum_{i=1}^{n}(g(x_i) - g(x_{i-1})) + \sum_{i=1}^{n}(h(x_i) - h(x_{i-1}))$, and the two sums telescope to $(g(b) - g(a)) + (h(b) - h(a))$. This bound is partition-independent, so taking the supremum over all partitions gives $V_a^b(f) \le (g(b) - g(a)) + (h(b) - h(a)) < \infty$ — $f$ is of bounded variation.

A sum of increasing functions is increasing: if $x_1 < x_2$ then $g(x_1) \le g(x_2)$ and $u(x_1) \le u(x_2)$, so $(g+u)(x_1) \le (g+u)(x_2)$ — and the same for $h + u$.

Then $(g + u) - (h + u) = g - h = f$ — the $u$ cancels.

Since there are infinitely many increasing functions $u$ (e.g. $u(x) = x$, $u(x) = x^2$, $u(x) = 2$ constant), there are infinitely many Jordan decompositions.

Uniqueness would need an extra normalization (e.g. $g(a) = 0$).