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Let $f:[a,b]\rightarrow \mathbb{R}$ be Riemann-integrable. I want to show, that $|f|^p$ is riemann-integrable if $p\in [1,\infty)$

The proof starts like this:

It suffices to show the claim for $0 \leq f \leq 1 $ (why? No further comment is given in the proof ) Then for every epsilon > 0 there exist step functions such that $$ 0 \leq \phi \leq f \leq \psi \leq 1$$ Then $\phi^p$ and $\psi^p$ are step functions for which $\phi^p \leq f^p \leq \psi^p$ and the mean value theorem tells us: $$ \phi^p - \psi^p \leq (\phi-\psi)p$$

How does the MVT come into play here? Nowhere in that equality I even see the differential of the function

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For $F(t)=t^p$, $t\in[0,1]$, the MVT says that there exists $\xi$ with $0\le\phi\le\xi\le\psi\le 1$ such that $$ \frac{F(\psi)-F(\phi)}{\psi-\phi}=F'(\xi)=p\xi^{p-1}\le p. $$ It gives the inequality (where you seem to interchange $\phi$ and $\psi$).

It is enough to consider $0\le f\le 1$ because $f$ is integrable iff $\frac{f-f_\min}{f_\max-f_\min}=C\cdot f+D$ is integrable, and the latter is between $0$ and $1$.

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  • $\begingroup$ Crystal clear and short! This is what should have been in the book in my opinion. Thank you very much! $\endgroup$ – Jonathan Apr 3 '17 at 16:22

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