0
$\begingroup$

I was wondering if there is a good general technique for evaluating the Riemann-Stieltjes integral in cases where the integrator is discontinuous on a few points. For example: find $$\int_{0}^{2} f(x) d\alpha(x)$$ where $f(x) = e^{2x}$ and $\alpha = x + 1$ for $x$ in $[0, 1]$ and $\alpha = 6x$ for $x$ in $(1, 2]$. (Jumps from 2 to 6.)

It should be Riemann integrable based on the Riemann condition. Then I would like to split it apart and then use the reduction to a Riemann integral in terms of $\alpha'(x)$, but the condition for this is that $\alpha'(x)$ is continuous, but the derivative does not even exist for the interval $[0, 2]$. So instead I thought of splitting the integral apart into $\int_{0}^{1}f(x)d\alpha(x) + \int_{1}^{2}f(x)d\alpha(x)$, where the first term exists. Then for the second we might take $\int_{1}^{1 + \epsilon} f(x)d\alpha(x) + \int_{1 + \epsilon}^{2}f(x)d\alpha(x)$. But I am not sure that taking the limit here as $\epsilon \to 0$ is relevant, since the integral as a function of the endpoints might not be continuous.

$\endgroup$

1 Answer 1

0
$\begingroup$

Your example is $\alpha(dx)=4\delta _1(dx)+g(x)dx$ where $g(x)=1$ on $[0,1]$ and $6$ on $(1,2]$ meaning that if $f$ is continuous on $[0,2]$ we have $$\int_{[0,2]}fd\alpha=4f(1)+\int_0^2f(x)g(x)dx$$ Discontinuities of the increasing function $\alpha$ correspond to Dirac masses of the corresponding positive measure $d\alpha(x)$ also denoted $\alpha(dx)$ in measure theory. It is a pity that many classes on measure theory do not stress the case of a measure on $R$ or an interval, essentially described by an increasing function.

$\endgroup$
1
  • $\begingroup$ Your integral is of the Lebesgue-Stieltjes type. As Martin R mentioned earlier, the Riemann-Stiletjes integral may not exists when the integrand $f$ and the integrator $g$ have common discontinuity of the same type (right or left). This is not the case with Lebesgue-Stieltjes since the integrator $g$ is right-continuous with left limits and becomes a (finite) measure. $\endgroup$
    – Mittens
    Jun 1, 2023 at 18:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .