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Do there exist $f$ and $g$ continuous functions on $[1,2]$ such that $\int f dg$ does not exist ?

I tried with

$f(x)=1$

$g(x) = \begin{cases} (x-1)\sin\left(\frac{1}{x-1}\right) & \text{if } x \neq 1 \\ 0 & \text{if } x = 1 \end{cases}$

I chose $g$ which is not a bounded variation function.

Is it good example?

Another example is :

$f(x)= \begin{cases} (x-1)\sin\left(\frac{1}{x-1}\right) & \text{if } x \neq 1 \\ 0 & \text{if } x = 1 \end{cases}$

$g(x) = \begin{cases} (x-1)\sin\left(\frac{1}{x-1}\right) & \text{if } x \neq 1 \\ 0 & \text{if } x = 1 \end{cases}$

but how can I prove it ?

Riemann-Stieltjes integral is really hard, so proving existence or non existence is hard.

Angelo
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Math Admiral
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    Mahmoud, $\int_1^2!f\mathrm dg$ exists by using the functions you chose, indeed in the first case it results that $$\int_1^2!!f,\mathrm dg=\int_1^2!!1,\mathrm dg=g(2)!-!g(1)=g(2)=\sin1;;,$$whereas in the second case ,$$\int_1^2!!f,\mathrm dg=\int_1^2!!g,\mathrm dg=\frac12\left[g(2)^2!-!g(1)^2\right]=\frac12g(2)^2=\frac12\sin^2!1,.$$ – Angelo Aug 03 '23 at 06:25
  • Is g differentiable at $1$? I recall that it isn't differentiable at $1$ – Math Admiral Aug 03 '23 at 06:34
  • Sorry, it exists but not continuous at $1$ – Math Admiral Aug 03 '23 at 07:16
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    If you know a continuous function that is not of bounded variation on $[1,2]$, then isn't the integral $\int_1^2 1dg$ (i.e. the choice $f=1$) not defined? You may only need one such function, then. – Sarvesh Ravichandran Iyer Aug 03 '23 at 07:21
  • @SarveshRavichandranIyer But doesn't $g'$ exists and continuous on $(1,2)$ and so $\int fdg = \int fg' dx$? (g is not of bounded variation) – Math Admiral Aug 03 '23 at 07:23
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    @Mahmoudalbahar If $g$ is not of bounded variation, then there is no possibility of $g'$ existing and being continuous on $(1,2)$. For example, read this answer and change any such function $f : [0,1] \to \mathbb R$ to $g :[1,2] \to \mathbb R$ by setting $g(x)=f(x-1)$. Then, $\int_1^2 1 dg$ won't exist. – Sarvesh Ravichandran Iyer Aug 03 '23 at 07:50

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