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In many theorems about the Riemann-Stieltjes integral they required the hypothesis of $f$ to be bounded to then conclude that $f$ is Riemann-Stieltjes integrable.

For example, suppose that $f$ is bounded in $I = [a,b]$, $f$ has only finitely many points of discontinuity in $I$, and that the monotonically increasing function $\alpha$ is continuous at each point of discontinuity of $f$, then $f$ is Riemann-Stieltjes integrable.

What if we remove the bounded hypothesis?

Could there exist an unbounded function $f$ in a given interval $[a,b]$ such that $\int_a^bf\,d\alpha$ exist?

Maybe a counterexample?

RRL
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user441848
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3 Answers3

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A function $f$ cannot be both unbounded and Riemann-Stieltjes integrable.

This can be shown by producing an $\epsilon > 0$ such that for any real number $A$ and any $\delta > 0$ there is a tagged partition $P$ with $\|P\| < \delta$ and with a Riemann-Stieltjes sum satisfying

$$|S(P,f,\alpha) - A| > \epsilon$$

Given any partition $P$, since $f$ is unbounded, it must be unbounded on at least one subinterval $[x_{j-1},x_j]$ of P. Using the reverse triangle inequality we have

$$|S(P,f,\alpha) - A| = \left|f(t_j)(\alpha(x_j) - \alpha(x_{j-1})) + \sum_{k \neq j}f(t_k)(\alpha(x_k) - \alpha(x_{k-1})) - A \right| \\ \geqslant |f(t_j)|(\alpha(x_j) - \alpha(x_{j-1})) - \left|\sum_{k \neq j}f(t_k)(\alpha(x_k) - \alpha(x_{k-1})) - A \right|$$

Since $f$ is unbounded on $[x_{j-1},x_j]$, choose a partition tag $t_j$ such that

$$|f(t_j)| > \frac{\epsilon + \left|\sum_{k \neq j}f(t_k)(\alpha(x_k) - \alpha(x_{k-1})) - A \right|}{\alpha(x_j) - \alpha(x_{j-1})},$$

and it follows that no matter how fine the partition $P$ we have

$$|S(P,f, \alpha) - A| > \epsilon.$$

Thus, when $f$ is unbounded, it is impossible to find $A$ such that for every $\epsilon > 0$ and sufficiently fine partitions, the condition $|S(P,f,\alpha) - A| < \epsilon$ holds. We can always select the tags so that the inequality is violated.

RRL
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  • if $f$ were bounded in $[x_{j-1},x_j]$, would it be impossible to have $\vert f(t_j)\vert >\frac{\epsilon +...}{...}$ ? because bounded means to have something like $\vert f\vert \le M ?$ – user441848 Jun 05 '17 at 20:14
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    Yes -- that is correct. If $f$ unbounded on $I$ we can for any $M$ no matter how large find a point $t_j$ such that $|f(t_j)| > M$. This, of course, violates the restrictive definition of the Riemann-Stieltjes integral where sums can be made close to the value of the integral no matter how the intermediate points (tags) are selected. – RRL Jun 05 '17 at 20:30
  • this is a side question but closely related: I suppose that we can extend the concept of improper integral to the Riemann-Stieltjes as we do with the integral of Riemann. If Im not wrong in this context we can (sometimes) evaluate integrals of unbounded functions, right? –  Jun 05 '17 at 22:14
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    @Masacroso: Sure -- you can always define an improper Riemann-Stieltjes integral in the usual way. For example $\int_a^\infty f , d\alpha = \lim_{c \to \infty} \int_a^c f , d\alpha$ if the limit exists. This was pointed out by Chappers. Where things are more complicated than the Riemann case occurs in integrals on bounded intervals where integrand $f$ and integrator $\alpha$ share a common point of discontinuity in the interval. In that case the R-S integral fails to exist regardless of the behavior everywhere else, (eg, only a single discontinuity point). – RRL Jun 05 '17 at 22:34
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    But the very strict definition of the Riemann or Riemann-Stieltjes integral prohibits unbounded functions. An unbounded function can never satisfy the requirements in that sense. If you are talking about the integral of an unbounded function, then it is in a broader context (i.e., Lebesgue) and an improper intetgral is just the consequence of a double limiting process. – RRL Jun 05 '17 at 22:39
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Remember that the Riemann/Darboux integral requires the function to be bounded, or at least one of the upper and lower sums for a given partition will always diverge. We see the same situation in the Darboux formulation of Riemann–Stieltjes integrability.

Of course, one can formulate an improper Riemann–Stieltjes integral in exactly the same way as the improper Riemann integral: see, e.g., Burkill & Burkill, § 6.3.

Chappers
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Great question. Great answers. Here's mine:

Assume that $g:[a,b]\to\mathbb{R}$ is not bounded and that $\alpha:[a,b]\to\mathbb{R}$ is strictly increasing. If we use the alternative definition of an integral, which I personally prefer, (that which uses upper and lower generalized Darboux sums, and which is presented as the definition of the integral in baby Rudin), then it will suffice to show that

$$ \exists\hspace{0.4mm} \varepsilon > 0 \ni \hspace{1mm}\forall P\in\mathcal{P}, \hspace{1mm} U(P,g,\alpha) \geq L(P,g,\alpha) + \varepsilon $$

where $\mathcal{P}$ denotes the set of all partitions of a given closed interval ${[a,b]}$. More to the point, we will prove the stronger statement that this holds for any $\varepsilon>0$. Fix an arbitrary $\varepsilon>0$ and an arbitrary partition $P := \{x_0,\ldots,x_n\}$ of ${[a,b]}$. Since $g$ is not bounded on ${[a,b]},$ we have that that $g$ is not bounded on $[x_{\ell-1},x_\ell]$ for some $\ell\in\{1,\ldots,n\}$. Consider this $\ell$. Since $\alpha$ is strictly increasing, $\alpha(x_\ell) - \alpha(x_{\ell-1}) := \Delta\alpha_\ell >0$. Then, since $g$ is not bounded on $[x_{\ell-1},x_\ell]$, it is reasonably clear that

\begin{align*} \sup_{x\in[x_{\ell-1},x_\ell]} g(x) \hspace{4mm}\geq\hspace{4mm} \frac{\varepsilon}{\Delta\alpha_\ell} \hspace{2mm}+ \inf_{x\in[x_{\ell-1},x_\ell]} g(x). \end{align*}

This can be made less terse, but we skip the formality. Therefore,

\begin{align*} U(P,g,\alpha) - L(P,g,\alpha) \hspace{1mm}&:=\hspace{1mm} \sum_i \underbrace{ \Bigg(\sup_{x\in[x_{i-1},x_i]} g(x) \hspace{2mm} - \inf_{x\in[x_{i-1},x_i]} g(x) \bigg) \Delta\alpha_i }_{\hspace{1mm}\geq\hspace{1mm} 0} \\ \hspace{1mm}&\geq\hspace{1mm} \underbrace{ \Bigg(\sup_{x\in[x_{\ell-1},x_\ell]} g(x) \hspace{2mm} - \inf_{x\in[x_{\ell-1},x_\ell]} g(x) \bigg) }_{ \hspace{1mm}\geq\hspace{1mm} \varepsilon/(\Delta\alpha_\ell) } \Delta\alpha_\ell \hspace{1mm}\geq\hspace{1mm} \varepsilon. \end{align*}

Thomas Winckelman
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