Given $f$ increasing on $[a,b]$, $g(x)\in R(\alpha)$ on $[a,b]$, $\alpha \in C([a,b])$ and $\alpha \in BV([a,b])$ $$ \beta(x)=\int_a^xg(z)d\alpha(z) \text{ on [a,b]} $$

Why is the additional assumption $f$ continuous and what theorem if any is needed to show the we can substitute $d\beta$ in the following? $$ \int_a^bfd\beta= \int_a^bfgd\alpha $$

I thought that by the definition of $\beta$ it is differentiable and continuous on $[a,b]$ and I just did a direct substitution $d\beta=g(x)d\alpha(x)$. With an unused assumption I am concerned I overlooked something.

  • $\begingroup$ It says f is increasing! $\endgroup$ – Mhenni Benghorbal Feb 7 '16 at 5:12
  • $\begingroup$ thanks, but why do we care about that and $f$ continuous when working with $d\beta$? $\endgroup$ – StackAbstraction Feb 7 '16 at 13:48
  • $\begingroup$ I've already shown that $f \in BV([a,b])$ and $\beta$ continuous so $f \in R(\beta)$. $\endgroup$ – StackAbstraction Feb 7 '16 at 13:55

In general, for any increasing $\beta$ on $[a,b]$, $$ \beta(b) - \beta(a) \geq \int_a^b \beta'(t) dt $$ $\beta'$ exists a.e. since $\beta$ is increasing, so $\beta'$ is Riemann integrable. The above turns into equality when $\beta$ is absolutely continuous.

  • $\begingroup$ Why do we care about $f$ when we are working with $d\beta$? This does not address $f$ is continuous. We know by definition of $\beta$ that it is differentiable and continuous right? $\endgroup$ – StackAbstraction Feb 7 '16 at 13:46
  • $\begingroup$ @stackabstraction By $f$ I meant $\beta$ $\endgroup$ – Henricus V. Feb 7 '16 at 14:30

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