I'm working on a problem, stated as follows:

If $f$ is integrable with respect to every bounded, monotone increasing function $g$ on $[a, b]$, then is $f$ continuous on $[a, b]$?

I have proved that the converse is true, I want to know whether this statement is correct. This is my work until now. Forgive me for my little work.

Since $g$ is monotone, increasing, Riemann's condition holds, so the term $$\sum_{j=1}^p (M_j-m_j)(g(x_j)-g(x_{j-1}))$$

could be made arbitrarily small for a partition $\pi=\{x_0, x_1,...x_p\} \in \Pi[a, b]$. The choice of $g$ is also arbitrary, so the difference $g(x_j)-g(x_{j-1})$ could also be made arbitrarily small. This would imply that the difference between $M_j$ and $m_j$ could be arbitrarily small. So I suspect that $f$ should be uniformly continuous. However, I don't know whether this logic is fine, and how to formally write this proof down. Thanks in advance.

  • $\begingroup$ Can you guess what happens to the Riemann sums if $f=g=$Heaviside step function on $[-1, 1]$? Or, in the same fashion, if I give you a not continuous function $f$, can you cook a $g$ that makes those discontinuity points "shine"? $\endgroup$
    – Uskebasi
    Sep 21, 2021 at 9:45
  • $\begingroup$ @Uskebasi I know that I can always make a function $g$ such that $f$ is integrable w.r.t $g$, but in this case, the choice of $g$ is arbitrary. $\endgroup$
    – Joshua Woo
    Sep 21, 2021 at 9:48

1 Answer 1


Let us assume that $a < c < b$ and show that $f$ is continuous at $c$. (The cases $c=a$ and $c=b$ require only minor modifications.)

Let $g:[a, b]\to \Bbb R$ be defined as $g(x) = 0$ for $x < c$ and $g(x) = 1$ for $x \ge c$. The existence of $A = \int_a^b f(x) dg(x)$ implies that for every $\epsilon > 0$ there is a $\delta > 0$ such that $$ \left| \left(\sum_{k=1}^n f(c_j) (g(x_j)-g(x_{j-1})\right) - A\right| < \epsilon $$ for every partition $P = (x_0, \ldots, x_n)$ of $[a, b]$ with $\operatorname{norm}(P) < \delta$ and every choice of $c_j \in [x_{j-1}, x_j]$.

If we choose the partition such that $ [x_{j-1}, x_j] = [c - \delta/3, c+\delta/3]$ for some $j$ then the Riemann-Stieltjes sum reduces to $f(c_j)$, and it follows that $$ |f(x) - A | < \epsilon \quad \text{for all } x \in [c - \delta/3, c+\delta/3] $$ which implies that $$ |f(x) - f(c) | < 2\epsilon \quad \text{for all } x \in [c - \delta/3, c+\delta/3] \, . $$

This shows that $f$ is continuous. (The uniform continuity follows from the general fact that a continuous function on a compact interval is uniformly continuous.)


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