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 $$-\left|\sum_{i=1}^nf(t_i)-f(t'_i)\right|\le-\sum_{i=1}^n \left|f(t_i)-f(t'_i)\right|\le -\sum_{i=1}^n -\left|f(t_i)\right|+\left|f(t'_i)\right|=\sum_{i=1}^n \left|f(t_i)\right|-\left|f(t'_i)\right| $$
