# Integrable w.r.t every bounded monotone increasing function, then continuous

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.

• 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"? Sep 21, 2021 at 9:45
• @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. Sep 21, 2021 at 9:48

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.)