What is the difference between Riemann and Riemann-Stieltjes integrals?

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.

Thanks.

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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