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Suppose that $\alpha \in BV[a,b]$ and $V(x)$ is its total variation function on $[a,b]$.Show that if $f\in R(\alpha)$,then $\mid f \mid \in R(V)$.

What I know so far is if $\alpha \in BV[a,b]$,then $\alpha=\alpha_1 - \alpha_2$.$\alpha_1$ and $\alpha_2$ are both increasing.Since $f \in R(\alpha)$,then,$f \in R(\alpha_1 + \alpha_2)$.My next step would be showing $V\leq\alpha_1+\alpha_2$.Is my idea correct?Can someone help me finish the rest part of proof?

Paramanand Singh
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Zetton
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  • Who is Remiann? – Karl Oct 14 '18 at 06:00
  • Bad idea to use both $\alpha$ and $a$ in a question. Anyway, $\alpha$ is a function – what does it mean for $f$ to be integrable over the function $\alpha$? And what does it mean to be integrable over $V$? – Gerry Myerson Oct 14 '18 at 06:54
  • You should first prove that $f\in R(V) $. This proof is difficult and long and available at https://math.stackexchange.com/a/2592615/72031 Further since $V$ is increasing we have the implication $f\in R(V) \implies |f|\in R(V) $ via triangle inequality $||p|-|q||\leq |p-q|$. – Paramanand Singh Oct 14 '18 at 09:03
  • @Paramanand Singh Thank you! – Zetton Oct 14 '18 at 18:39
  • @Paramanand Singh:I checked your link.While in the last step,he considered two subintervals where $g$ is increasing or decreasing.But it seems like he only considered where $g$ is increasing in the following proof. – Zetton Oct 14 '18 at 21:08
  • Also see https://math.stackexchange.com/a/2033164/72031 – Paramanand Singh Oct 14 '18 at 22:31

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