How Are Borel Measures Related to Functions of Bounded Variation?

A function of bounded variation and a Borel measure are different kinds of objects, but on the real line they are closely connected. The key link is that changes in a right-continuous function of bounded variation can be recorded as the masses of intervals. This produces a finite signed Borel measure, often called the function’s Lebesgue–Stieltjes measure.

The converse works too: the cumulative mass of a finite signed Borel measure is a right-continuous function of bounded variation. The choice of representative matters at jump points, so the standard convention is to use right-continuous functions and half-open intervals.

From bounded variation to a measure

A real-valued function $F$ has bounded variation on an interval if the sums of its absolute increments are bounded over all finite ordered partitions. On the real line, finite total variation means

$$\sup_{x_0<\cdots<x_n}\sum_{i=1}^{n}|F(x_i)-F(x_{i-1})|<\infty.$$

Suppose $F$ is right-continuous and has finite total variation on $\mathbb{R}$. Define interval increments by

$$\mu((s,t])=F(t)-F(s),\qquad s<t.$$

These increments extend to a finite signed Borel measure $\mu$. The construction is the Lebesgue–Stieltjes construction. One way to understand why it works is to write $F$ as a difference of two bounded, nondecreasing, right-continuous functions. Each nondecreasing function defines a positive Lebesgue–Stieltjes measure; subtracting the two measures gives $\mu$.

If $F$ is only defined on a compact interval $[a,b]$, the same formula defines a measure on $(a,b]$. A mass at $a$ is not determined by increments inside the interval, so specifying a measure on all of $[a,b]$ requires an additional endpoint convention.

From a measure to a function

Let $\mu$ be a finite signed Borel measure on $\mathbb{R}$. Its cumulative function is

$$F(x)=\mu((-\infty,x]).$$

For $s<t$, this gives

$$F(t)-F(s)=\mu((s,t]).$$

The function is right-continuous: as $x_n$ decreases to $x$, the sets $(-\infty,x_n]$ decrease to $(-\infty,x]$, and a finite measure is continuous under decreasing intersections. For signed measures, apply this fact to the positive and negative parts.

For any increasing finite sequence $x_0<\cdots<x_n$,

$$\begin{aligned} \sum_{i=1}^{n}|F(x_i)-F(x_{i-1})| &=\sum_{i=1}^{n}|\mu((x_{i-1},x_i])|\\ &\leq \sum_{i=1}^{n}|\mu|((x_{i-1},x_i])\\ &\leq |\mu|(\mathbb{R}), \end{aligned}$$

where $|\mu|$ is the total variation measure. Thus $F$ has bounded variation. Adding a constant to $F$ does not change its associated measure, because the measure records increments, not the function’s absolute level.

Worked example: a jump becomes an atom

Let

$$F(x)= \begin{cases} 0,&x<2,\\ 3,&x\geq 2. \end{cases}$$

This function is right-continuous and has total variation $3$. Its associated measure satisfies $\mu((s,t])=F(t)-F(s)$. If $s<2\leq t$, the increment is $3$; if the interval does not cross $2$, it is $0$. Therefore

$$\mu=3\delta_2,$$

where $\delta_2$ is the unit point mass at $2$. In particular, $\mu(\{2\})=3$. More generally, the mass at a point records the jump:

$$\mu(\{x\})=F(x)-F(x-),$$

where $F(x-)$ is the left-hand limit.

What changes for a function that is not right-continuous?

A function of bounded variation need not be right-continuous. Its values at jump points can be assigned in different ways without changing its one-sided limits, and those choices affect which interval-increment formula applies. The usual approach is to replace it with its right-continuous version, when that version is available, and then use half-open intervals $(s,t]$. Alternatively, one can choose a left-continuous convention and use a different choice of interval endpoints. State the convention before identifying jump sizes with atoms.

Common mistakes

• Treating the function itself as the measure. The measure is built from increments of the function; it is not the same object as $F$.
• Ignoring endpoint conventions. The formula on $(a,b]$ does not specify a possible atom at $a$.
• Assuming every BV function has a unique associated measure without a convention. Right-continuity and the interval convention determine how jumps are assigned.
• Confusing signed measures with positive measures. A general BV function can increase and decrease, so its associated measure may have positive and negative parts. If $F$ is nondecreasing, its associated measure is positive.