# Notation question, Stieltjes integral

I'm reading a signal processing textbook and they frequently write out formulas where if I have a Stieltjes integral:

$$\int_a^b f(t) \, dX(t)$$

They'll write an explicit formula for $dX(t)$, such as

$$dX(t) = g(t)$$

Is the best way to interpret this notation to read my integral as the limit of the following sum?

$$\sum f(t^*) (g(t_i) - g(t_{i-1}) )$$

The notation that I would expect to see would be to write $X(t) = g(t)$ and then interpret the integral as the limit of the above sum.

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Yes, you got it right. I think though you could do things even simpler and directly write $$\sum f(t*)\left(X(t_i)-X(t_{i-1})\right)$$ without resourcing to that $\,g(x)\,$ – DonAntonio Aug 16 '12 at 16:17
Are there any notable textbooks at a first-year grad. student level that employ that notation? I'm used to seeing $dX$ reserved for the increment, $X(t_i) - X(t_{i-1})$ – ncRubert Aug 16 '12 at 16:22
Most decent analysis books even at undergraduate level, deal with Riemann-Stieltjes integral. For example, Lang's "Undergraduate Analysis" , Bruckner et al.'s "Real Analysis", Rudin's "Principles of Mathematical Analysis", etc. – DonAntonio Aug 16 '12 at 16:32
What you say you would expect is closer to the truth than is what you said you read. But there's no need to use two letters, $X$ and $g$ to refer to the same function. One is enough. – Michael Hardy Aug 16 '12 at 17:27

"$dX(t)=g(x)$" makes no sense. One can write $dX(t)=g(t)\,dt$, with $t$ rather than $x$ as the variable, and with "$dt$" after it, provided $X$ is differentiable and $X'=g$. But the Riemann-Stieltjes integral makes sense even in many cases where $X$ is not differentiable, and is the limit of the sum you wrote provided you put $X(t_i)-X(t_{i-1})$ where you had $g(t_i)-g(t_{i-1})$.
• "$dX(t) = g(t)$" doesn't make sense without "$dt$", i.e. one has $dX(t)=g(t)\,dt$ provided $X'=g$. But the Riemann-Stieltjes integral can make sense even in many cases where $X$ is not differentiable.
• You need the limit of the sum with $X(t_i)-X(t_{i-1})$, not with $g(t_i)-g(t_{i-1})$.