Integrator monotonically increasing ? R.S. Integral

[sihag]

Could someone explain why the R.S. Integral is defined for a monotonically increasing integrator? Can't we use a decreasing fuction anologously?

 

[HallsofIvy]

Yes, you could. It would just give the negative of the corresponding integral with an increasing integrator.

(For anyone who is wondering, "R.S" is the Riemann-Stieljes integral. It is defined exactly like the Riemann integral except that instead of measuring the size of each interval forming the base of a rectangle as [itex]x_{i+1}- x_i[/itex], we use [itex]\alpha(x_{i+1})-\alpha(x_i)[/itex] where [itex]\alpha(x)[/itex] can be an monotone increasing function of x. It is typically written [itex]\int f(x)d\alpha(x)[/itex]. If [itex]\alpha[/itex] is a differentiable function, the Riemann-Stieljes integral is exactly the same as the Riemann integral [itex]\int f(x) d\alpha/dx dx[/itex]. The interesting situation is when [itex]\alpha[/itex] is not differentiable. In particular, if [itex]\alpha[/itex] is the unit step function, and a and b are integers, then [itex]\int_a^b f(x) d\alpha= f(a)+ f(a+1)+ \cdot\cdot\cdot+ f(b)[/itex].)