Lebesgue–Stieltjes integration

Remarks

Example: let $g = \mathbf{1}_{[0, \infty)}$, the Heaviside step function with $g(0) = 1$, which is non-decreasing and right-continuous. Then $\mu_g$ is the Dirac measure at $0$, so $\int f \, dg = f(0)$ for every Borel function $f$. For $g(x) = x$ the measure $\mu_g$ is Lebesgue measure, and the integral is the usual Lebesgue integral.

Non-example: $g(x) = -x$ is not non-decreasing, so the definition does not apply directly. The formula $\mu_g((a, b]) = g(b) - g(a)$ would give negative mass.

For $g$ of bounded variation and right-continuous, one writes $g = g_1 - g_2$ with $g_1, g_2$ non-decreasing and right-continuous, and sets $\int f \, dg = \int f \, dg_1 - \int f \, dg_2$. On $[a, b]$ one writes $\int_{(a, b]} f \, dg$, and the endpoints matter when $g$ jumps there. If $f$ is continuous on $[a, b]$, the integral over $(a, b]$ equals the Riemann–Stieltjes integral $\int_a^b f \, dg$. If $g$ is absolutely continuous on bounded intervals, then $\int f \, dg = \int f(x) g'(x) \, dx$.