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Let $f:[a,b]\to \mathbb{R}$ be integrable. Prove $\left|f\right|$ is integrable with the definition of the integral using the Riemann sums (not the Darboux one).

I think the way to go is to use the following two theorems:

Cauchy criterion: A function $f:[a,b]\to \mathbb{R}$ is integrable if and only if $\forall \epsilon>0\exists \delta>0$ so that for any pair of tagged partitions $\dot{\mathcal{P}},\dot{\mathcal{Q}}$ of $[a,b]$ \begin{equation}\left\|\mathcal{P}\right\|,\left\|\mathcal{Q}\right\|<\delta\implies \left|S(f,\dot{\mathcal{P}})-S(f,\dot{\mathcal{Q}})\right|<\epsilon\end{equation}

or the simpler squeeze theorem: $f:[a,b]\to \mathbb{R}$ is integrable if and only if $\forall \epsilon>0$ there exist integrable functions $g,h:[a,b]\to \mathbb{R}$ so that \begin{equation}g\le f\le h\text{ and }\int_a^bh-g<\epsilon\end{equation}

When I tried the Cauchy Criterion, I couldn't prove \begin{equation}\left|S(\left|f\right|,\dot{\mathcal{P}})-S(\left|f\right|,\dot{\mathcal{Q}})\right|<\left|S(f,\dot{\mathcal{P}})-S(f,\dot{\mathcal{Q}})\right|\end{equation} which doesn't really seem to be true. With the squeeze theorem I can't construct $h$. Any ideas?

EDIT: $$\left|S(\left|f\right|,\dot{\mathcal{P}})-S(\left|f\right|,\dot{\mathcal{Q}})\right|\le\left|S(f,\dot{\mathcal{P}})-S(f,\dot{\mathcal{Q}})\right|\iff \\ \left|\sum_{i=1}^n(\left|f(t_i)\right|-\left|f(t'_i)\right|)(x_i-x_{i-1})\right|\le \left|\sum_{i=1}^n(f(t_i)-f(t'_i))(x_i-x_{i-1})\right|\iff\\ -\left|\sum_{i=1}^nf(t_i)-f(t'_i)\right|\le \sum_{i=1}^n\left|f(t_i)\right|-\left|f(t'_i)\right|\le \left|\sum_{i=1}^nf(t_i)-f(t'_i)\right| $$ Now, $$\left|\sum_{i=1}^nf(t_i)-f(t'_i)\right|\ge\sum_{i=1}^n \left|f(t_i)-f(t'_i)\right|\ge \sum_{i=1}^n \left|f(t_i)\right|-\left|f(t'_i)\right| $$ and $$-\left|\sum_{i=1}^nf(t_i)-f(t'_i)\right|\le-\sum_{i=1}^n \left|f(t_i)-f(t'_i)\right|\le -\sum_{i=1}^n -\left|f(t_i)\right|+\left|f(t'_i)\right|=\sum_{i=1}^n \left|f(t_i)\right|-\left|f(t'_i)\right| $$

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Work more on $\left|S(\left|f\right|,\dot{\mathcal{P}})-S(\left|f\right|,\dot{\mathcal{Q}})\r‌​ight|<\left|S(f,\dot{\mathcal{P}})-S(f,\dot{\mathcal{Q}})\right|$ but of course only ${}\le{}$ is the goal. –  GEdgar Jan 22 '13 at 13:18
    
@GEdgar It seems you were right. Could you please check my solution in the edited question? How is the boundedness of $f$ used? –  Optional Jan 22 '13 at 13:31
    
Related question (perhaps a duplicate): math.stackexchange.com/questions/316090/… –  Martin Sleziak Jan 20 at 15:20

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