In our functional analysis class we quickly introduced the Riemann-Stieltjes-Integral as follows:
If $g$ is a function of bounded variation on $[a,b]$ and $f$ is a step function with values $c_i$ in the intervals $(t_{i-1},t_i)$, where $\{t_0,...,t_N\}$ is a partition of $[a,b]$, then its Riemann-Stieltjes-Integral is defined as \begin{equation}\int_a^bf(t)dg(t):=\sum_{i=1}^Nc_i(g(t_i)-g(t_{i-1})). \end{equation} As $g$ is of bounded variation, this can easily seen to be continuous on the space $SF[a,b]$ of step functions on the interval [a,b], so it has a unique extension to $cl(SF[a,b])^{|| • ||_{\infty}}\supset C([a,b])$. This extension is also called Riemann-Stieltjes-Integral.
On the other hand, I‘m familiar with the concept of a measure with density, i.e. if $\mu$ is some measure, then \begin{equation}\lambda(A):=\int \chi_A g d\mu \end{equation} is also a measure on the same sigma-algebra (assuming $g$ is non-negative and measurable).
I was wondering how these are related. In which cases are these the same? I know from basic measure theory that \begin{equation}\int f d\lambda=\int f g d\mu. \end{equation}So can we then say \begin{equation}\int_a^b f(t)dg(t)=\int_a^b f(t)g(t)dt? \end{equation}
If this is correct, I‘d be glad for outlines or ideas of proof (or on which other results they relay on).
Many thanks in advance.
No. Your formula is wrong. What is true is if $g$ is smooth (and $g'$ is integrable) then $\int_a^{b}f(t)dg(t)=\int_a^{b} f(t)g'(t)dt$.