It seems to me that the standard definition of the Riemann-Stieltjes integral is too strict as to almost never exist.

First, I want to recall the definition of the Riemann-Stieltjes integral, which I am taking from Wheeden-Zygmund page 23. $\int_a^b f d\phi = S$ is defined to mean $\displaystyle{\lim_{\|\Gamma \| \rightarrow 0} \sum_{\Gamma} f d\phi } = S$, where $\Gamma = \{ (t_i)_{i=1}^N, (x_i)_{i=0}^N\} $ is a tagged partition of $[a, b]$ and $\displaystyle{\sum_{\Gamma} f d\phi} = \displaystyle{\sum_{i=0}^N f(t_i)(\phi(x_{i+1}) - \phi(x{i}))}$ and $\|\Gamma \| = \text{min}\{ x_{i+1} - x_{i} \}_{i = 0}^{N-1}$.

Now, let's take the standard example of a probability mass function, which can also be found on page 23. Let $\phi$ be a step function that is constant on $(\alpha_{i-1}, \alpha_i)$ for $i = 1, \ldots, m$ where $a = \alpha_0 < \alpha_1 < \ldots < \alpha_m = b$ for some $\alpha_i's$. Let $d_i = \phi(\alpha_i+) - \phi(\alpha_i-)$ for $i = 1, \ldots, m-1$, where +, - denote the right and left hand limits, and let $d_0 = \phi(\alpha_0+) - \phi(\alpha_0)$ and $d_m = \phi(\alpha_m) - \phi(\alpha_m-)$. Now, consider a sequence $\Gamma_k$ of tagged partitions such that $\| \Gamma_k \| < \frac{1}{k}$ and each $\alpha_i$ are endpoints of some interval in $\Gamma_k$ for each $k$. It follows that $\displaystyle{\sum_{\Gamma_k} f d\phi } = 0$ for every $k$ since the $d\phi = 0 $ on each interval of each $\Gamma_k$. Let $\Gamma_k'$ be another sequence of tagged partitions such that $\| \Gamma_k' \| < \frac{1}{k}$ for each $k$, and each $\alpha_i$ is included in intervals in $\Gamma_k$ with tag $f(\alpha_i)$. It then follows that $\displaystyle{\sum_{\Gamma_k'} f d\phi } = \sum_{i=0}^m f(\alpha_i) d_i$ for every $k$.

The existence of the Riemann-Stieltjes integral implies that for $\textit{any }$ sequence of tagged partitions $(\Gamma_k)_{k=0}^\infty$ such that $\|\Gamma \| \rightarrow 0$, $\displaystyle{\sum_{\Gamma_k} f d\phi } \rightarrow S$. However, we see that $\displaystyle{\sum_{\Gamma_k} f d\phi } \rightarrow 0$ and $\displaystyle{\sum_{\Gamma_k'} f d\phi } \rightarrow \sum_{i=0}^m f(\alpha_i) d_i$, so the Riemann-Stieltjes integral does not exist in this case. This is, of course, the standard case of a probability mass function, so it should indeed be integrable.

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    $\begingroup$ No! Riemann-Stieltjes is defined on, and using, closed intervals, so your "end-point" $x_i$ will have to be evaluated instead of only the left- and right-hand limits $\phi(x_i-)-\phi(x_{i-1}+)$. Indeed the R-S integral will pick up the jump discontinuities of $\phi$. $\endgroup$ Aug 1, 2020 at 22:42
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    $\begingroup$ It seems like you are claiming that $\phi(x_{i+1})-\phi(x_i) = 0$ for your choice of tagged partition $\Gamma$, but this is not true. For instance, if $f\equiv 1$, then $$\sum_{i=0}^{N-1}f(t_i)(\phi(x_{i+1})-\phi(x_i))=\sum_{i=0}^{N-1}(\phi(x_{i+1})-\phi(x_i))=\phi(b)-\phi(a).$$ As pointed out by the above comment, it seems that you are confusing $\phi(x_{i+1})-\phi(x_i)$ with the quantity $\phi(x_{i+1}^-)-\phi(x_i^+)$. (The latter quantity is indeed zero for your choice of tagged partitions.) $\endgroup$ Aug 2, 2020 at 0:08
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    $\begingroup$ On a more general note, the Riemann-Stieltjes integral can indeed not exist for specific probability mass functions, when we may 'feel' that the integral should exist. This is why the Lesbesgue integral was introduced that addresses that problem. $\endgroup$ Aug 2, 2020 at 8:14

1 Answer 1


This is a classical theorem that includes the classic Riemann integral existence theorem as a special case, which is: $\int_a^b fdx$ exists as a Riemann integral iff the set of discontinuities of $f$ is of $0$ measure.

Theorem (Riemann-Stieltjes Integral Existence): Let $f$ and $g$ be bounded real functions on $[a,b]$, and suppose that $g$ is non-decreasing on $[a,b]$. Then $f$ is Riemann-Stieltjes integrable with respect to $g$ iff the set of discontinuities of $f$ is of $g$-measure zero.

In particular, if $g$ has a jump discontinuity at $x \in [a,b]$, then $f$ cannot have a discontinuity at $x$ in order for the Riemann-Stieltjes integal $\int_a^b fdg$ to exist.


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