Chemistry - Riemann-Stieltjes integral

In mathematics, the Riemann-Stieltjes integral is a generalization of the Riemann integral. The Riemann-Stieltjes integral of a real-valued function f of a real variable with respect to a nondecreasing real function g is denoted by

$\int_a^b f(x) \, dg(x)$

and defined to be the limit as the mesh of the partition of the interval [a, b] approaches zero, of the sum

$\sum_{x_i\in P} f(c_i)(g(x_{i+1})-g(x_i))$

where c_i is in the ith subinterval [x_i, x_i+1]. In order that this Riemann-Stieltjes integral exist it is necessary that f and g do not share any points of discontinuity in common. The two functions f and g are respectively called the integrand and the integrator.

For another formulation of integration that is more general, see Lebesgue integration.

Contents

1 Properties and relation to the Riemann integral

2 What if g is not monotone?

3 Application to probability theory

1 See also

Properties and relation to the Riemann integral

If g should happen to be everywhere differentiable, then the integral is no different from the Riemann integral

$\int_a^b f(x) g'(x) \, dx.$

However, g may have jump discontinuities, or may have derivative zero almost everywhere while still being continuous and nonconstant (for example, g could be the Cantor function or the question mark function), in either of which cases the Riemann-Stieltjes integral is not captured by any expression involving derivatives of g.

The Riemann-Stieltjes integral admits integration by parts in the form

$\int_a^b f(x) \, dg(x)=f(b)g(b)-f(a)g(a)-\int_a^b g(x) \, df(x).$

What if g is not monotone?

Somewhat more generally, one may define a Riemann-Stieltjes integral with respect to any function g of bounded variation, since every such function can be written uniquely as a difference between two nondecreasing functions; the integral is the corresponding difference between two Riemann-Stieltjes integrals with respect to nondecreasing functions.

Application to probability theory

If g is the cumulative probability distribution function of a random variable X that has a probability density function with respect to Lebesgue measure, and f is any function for which the expected value E(|f(X)|) is finite, then, as is well-known to students of probability theory, the probability density function of X is the derivative of g and we have

$E(f(X))=\int_{-\infty}^\infty f(x)g'(x)\, dx.$

But this formula does not work if X does not have a probability density function with respect to Lebesgue measure. In particular, it does not work if the distribution of X is discrete (i.e., all of the probability is accounted for by point-masses), and even if the cumulative distribution function g is continuous, it does not work if g fails to be absolutely continuous (again, the Cantor function may serve as an example of this failure). But the identity

$E(f(X))=\int_{-\infty}^\infty f(x)\, dg(x)$

holds if g is any cumulative probability distribution function on the real line, no matter how ill-behaved.