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I'm quite confused, what is the difference between these two integrals (R and RS)? It seems that RS is closer to Lebesgue in its treatment of discontinuities, but otherwise I don't understand. If someone could give an example of a function for which they were different, it would be very beneficial.


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It seems to me that you are integrating relative to a $dg(x)$, rather than $dx$. For example, the if $g(x)$ is $0$ for negative $x$ and $1$ for positive $x$, then then $\int_{-1}^{1} f(x)dg(x)$ is $f(0)$ if $f$ is continuous.

If $g(x)=x$, the Riemann-Stieltjes integral is just the Riemann integral.

If $g(x)$ is continuously differentiable, then the RS-integral $\int_{a}^{b} f(x)dg(x)$ is the same as the Riemann integral $\int_a^b f(x)g'(x) dx$.

The differences then are the cases where $g(x)$ is not continuously differentiable. For example, if $g(x)$ is the step function above, then $dg(x)$ is "like" the Dirac delta function.

It's a beginning of a way of thinking of integrals as operators.

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For the case with Dirac delta function, doesn't the RS integral reduce to R = $\int_{0}^{1}f(x)dx$? –  user19821 Jan 25 '12 at 4:06
Nope, because if $g$ is the step function, and you partition the interval $-1=x_0<x_1<..<x_n=1$, the sum $\sum_i f(c_i)(g(x_{i+1})-g(x_i))$ is zero at every $i$ except where $x_i\leq 0 < x_{i+1}$, and, in that term, $g(x_{i+1})-g(x_i)$ is $1$, so the value of the sum is $f(c_i)$ where $c_i$ is in a neighborhood of $0$. That means that the limit will be $f(0)$ if $f$ is continuous at $0$. If $g(x)=0$ for $x\leq 0$ and $g(x)=x$ for $x>0$ then $\int_{-1}^1 f(x)dg(x)=\int_{0}{1} f(x)fx$. –  Thomas Andrews Jan 25 '12 at 4:31
did you mean $\int_{0}^{1}f(x)dx$? –  user19821 Jan 25 '12 at 6:06
Yes, got a typo in there. –  Thomas Andrews Jan 25 '12 at 7:53

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