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I got this question which I think I know the solution to but I am not certain I am doing it right, will gladly use some help about it.

So the question is like that, assume we have function $\alpha$ which is monotonic on $\left[a,b\right]$, and function f such that $\intop_{a}^{b}f\left(x\right)d\alpha\left(x\right)$ exists. I need to show that $\intop_{a}^{b}f^{2}\left(x\right)d\alpha\left(x\right)$ exists as well.

So what I have done till now is, assume $\alpha$ is increasing, let be $\varepsilon>0$ , there exists partition $P=\left\{ x_{i}\right\} _{i=0}^{n}$ such that: $$\left|\overline{S}_{f,\alpha}\left(P\right)-\underline{S}_{f,\alpha}\left(P\right)\right|<\varepsilon$$ so we get: $$\left|\overline{S}_{f,\alpha}\left(P\right)-\underline{S}_{f,\alpha}\left(P\right)\right|=\sum_{i=1}^{n}\sup_{x_{1},x_{2}\in\left[x_{i-1},x_{i}\right]}\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|\left(\alpha\left(x_{i}\right)-\alpha\left(x_{i-1}\right)\right)$$ and now my problem arise, I assume that if $\alpha$ is constant so the integral obviously exist, and else f must be bounded on $\left[a,b\right]$ (that's the thing I can't manage to prove). so we get: $$\sup_{x_{1},x_{2}\in\left[x_{i-1},x_{i}\right]}\left|f^{2}\left(x_{1}\right)-f^{2}\left(x_{2}\right)\right|=\sup_{x_{1},x_{2}\in\left[x_{i-1},x_{i}\right]}\left|f^{2}\left(x_{1}\right)-f\left(x_{1}\right)f\left(x_{2}\right)+f\left(x_{2}\right)f\left(x_{1}\right)-f\left(x_{2}\right)\right|\\\leq\sup_{x_{1},x_{2}\in\left[x_{i-1},x_{i}\right]}\left|f\left(x_{1}\right)\right|\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|+\left|f\left(x_{2}\right)\right|\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|\\\leq2M\sup_{x_{1},x_{2}\in\left[x_{i-1},x_{i}\right]}\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|$$ and we get: $$\left|\overline{S}_{f^{2},\alpha}\left(P\right)-\underline{S}_{f^{2},\alpha}\left(P\right)\right|=\sum_{i=1}^{n}\sup_{x_{1},x_{2}\in\left[x_{i-1},x_{i}\right]}\left|f^{2}\left(x_{1}\right)-f^{2}\left(x_{2}\right)\right|\left(\alpha\left(x_{i}\right)-\alpha\left(x_{i-1}\right)\right)\\\leq2M\sum_{i=1}^{n}\sup_{x_{1},x_{2}\in\left[x_{i-1},x_{i}\right]}\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|\left(\alpha\left(x_{i}\right)-\alpha\left(x_{i-1}\right)\right)\\<2M\varepsilon$$so we just choose the right $\varepsilon$ at the beginning and we done. But I cant manage to prove f is bounded. I know how to prove it if $\alpha$ is strictly increasing, but if I just assume it's not constant i can't manage to see it because there might be infinity intervals in which is $\alpha$ is strictly increasing, and in each interval f must be bounded, but it doesn't mean it is bounded on all $\left[a,b\right]$ right?

Any help would be welcomed :) and also is my proof is even correct?

Abzikro
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  • If a function is Riemann-Stieltjes integrable with respect to a monotonic integrator then it must be bounded. See this answer. – RRL Jun 16 '22 at 09:49
  • @RRL isn't that true only if integrator is strictly monotonic? that's my problem with this question. let's say in his answer $\alpha\left(x_{i}\right)-\alpha\left(x_{i-1}\right)=0$. so it won't work.

    I thought maybe taking all the subintervals in which $\alpha$ is strictly monotonic. or maybe the question is just in fault and they meant strictly monotonic...

     – Abzikro Jun 16 '22 at 11:23
  • So you are saying that an unbounded function can be Riemann-Stieltjes integrable with respect to an integrator that is not strictly monotone (e.g., non-decreasing but not strictly increasing). – RRL Jun 16 '22 at 14:56

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