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Assume that $f$ is Riemann integrable over $[a, b]$. Show that $|f|$ is Riemann integrable over $[a,b]$

(Note that the definition of Riemann integrability that I'm using is through Darboux sums)

Attempted proof: Let $\epsilon > o$ be given, then there exists a partition $P = \{x_i\}_{i = 0}^n$ of $[a, b]$ such that $U(f, P), -L(f, P) < \epsilon$.

Then we have $$U(|f|, P) = \sum_{i=1}^n \sup_{x \in [x_{i-1}, x_i]}|f(x)|(x_i - x_{i-1}) = \sum_{i=1}^n \left|\inf_{x \in [x_{i-1}, x_i]}f(x)\right|(x_i - x_{i-1})$$ which implies that $$U(|f|, P) = \sum_{i=1}^n |m_i|(x_i - x_{i-1})$$

and similarly we have $$L(|f|, P) = \sum_{i=1}^n \inf_{x \in [x_{i-1}, x_i]}|f(x)|(x_i - x_{i-1}) = \sum_{i=1}^n \left|\sup_{x \in [x_{i-1}, x_i]}f(x)\right|(x_i - x_{i-1})$$

which implies that $$L(|f|, P) = \sum_{i=1}^n |M_i|(x_i - x_{i-1})$$

Thus $U(|f|, P) - L(|f|, P) = \sum_{i=1}^n\left(|m_i| - |M_i|\right)(x_i - x_{i-1})$ and since for any $i \in \{0, ..., n\}$ we have $|m_i|< |M_i|$ it follows that $|m_i| - |M_i| < 0$ hence $U(|f|, P) - L(|f|, P) < 0 < \epsilon$ and by the criterion for integrability we have $|f|$ to be Riemann integrable over $[a, b]$. $\square$


Is the above proof correct? Have I made any mistakes or errors?

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  • $\begingroup$ What is your justification for $\sum_{i=1}^n \sup_{x \in [x_{i-1}, x_i]}|f(x)|(x_i - x_{i-1}) = \sum_{i=1}^n \left|\inf_{x \in [x_{i-1}, x_i]}f(x)\right|(x_i - x_{i-1})$? $\endgroup$ – Michael Biro May 13 '18 at 16:36
  • $\begingroup$ Take $f(x)=x$ on $[0,1]$. Then $\sup|f|=1$ while $|\inf f|=0$. Same problem with $\inf|f|$ and $|\sup f|$. $\endgroup$ – A.Γ. May 13 '18 at 16:42
  • $\begingroup$ A short proof using a lemma in the appendix of Stein's Fourier Analysis: $f \in \mathcal{R}[a,b] \iff$ set of discountinuous points of $f$ has measure zero, so ... $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 13 '18 at 17:00
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Hint:

$$\sup_{x \in I} |f(x)| - \inf_{x \in I} |f(x)| = \sup_{x,y \in I} |\,|f(x)| - |f(y)|\,| \leqslant \sup_{x,y \in I} |\,f(x) - f(y)\,| = \sup_{x \in I} f(x) - \inf_{x \in I} f(x)$$

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