(Verifying) Riemann Integral of $g(x)=x$ on $[0,1]$ using tagged partitions. I need to verify whether my method is correct or not in the following problem. 
Let $g(x)=x$ on $I=[0,1]$.

*

*Let
$
\dot{\mathcal{P}} = \left\{ \left( \left[ x_{i-1},x_i \right],t_i \right) \right\}_{i=1}^n
$
be a tagged partition of $I$, with $t_i = \frac{x_{i} + x_{i-1}}{2}$. Compute $S(g\,;\dot{\mathcal{P}})$.


*$\forall \, \varepsilon \gt 0$ and
$
\dot{\mathcal{Q}} = \left\{ \left( \left[ x_{i-1},x_i \right],q_i \right) \right\}_{i=1}^n
$
a tagged partition of $I$, with $\| \dot{\mathcal{Q}}\| \lt \varepsilon$ prove that
$$
\vert S(g\,;\dot{\mathcal{Q}}) - S(g\,;\dot{\mathcal{P}}) \vert
= \vert S(g\,;\dot{\mathcal{Q}}) - \frac{1}{2} \vert \lt \varepsilon.
$$


*Conclude that $g \in \mathcal{R}[0,1]$ and that $\int_{0}^{1} g = \frac{1}{2}$
 
My answer :




*

*

*

*$S(g\,;\dot{\mathcal{P}}) = \frac{1}{2}$ via the def. of the Riemann sum.

*$g \in \mathcal{R}[0,1]$ Iff $ \exists \, L \in \mathbb{R} ,\forall \, \varepsilon \gt 0, \exists \delta_{\varepsilon} \gt 0,\forall \, \dot{\mathcal{Q}} \text{ s.t. } \| \dot{\mathcal{Q}} \| \lt \delta_{\varepsilon} \Rightarrow
\vert S(g\,;\dot{\mathcal{Q}}) - L \vert \lt \varepsilon $

Here : $L = S(g\,;\dot{\mathcal{P}}) = \frac{1}{2} \,;\, \delta_{\varepsilon} = \varepsilon$

Giving : $\vert S(g\,;\dot{\mathcal{Q}}) - \frac{1}{2} \vert \lt \varepsilon $.

*From part (2) $g \in \mathcal{R}[0,1]$ and $L = \int_0^1 g = \frac{1}{2}$

Thank you in advance.

Edit : Thank you for correcting me, my apologies the $t_{i} = \frac{x_{i} - x_{i-1}}{2}$ was a typo. indeed.



 A: The choice $t_i = \frac{x_1-x_{i-1}}{2}$ is not a valid tag since it is not guaranteed that $t_i \in [x_{i-1},x_i]$ for all $i$. For example, given the partition points $\{0,\frac{1}{2},1\}$ we have $t_2 = \frac{1-\frac{1}{2}}{2} = \frac{1}{4} \notin \left[\frac{1}{2},1\right]$.  This is most likely a typographical error and the tag should be $t_i = \frac{x_i+x_{i-1}}{2}$, so that
$$S(g\,;\dot{\mathcal{P}})= \sum_{i=1}^n\frac{x_i+x_{i-1}}{2}(x_i - x_{i-1})= \frac{1}{2}\sum_{i=1}(x_i^2-x_{i-1}^2) = \frac{1}{2}(1-0) = \frac{1}{2}$$
This is the only tag choice that yields a constant value of $\frac{1}{2}$ for every Riemann sum regardless of the spacing and the number of partition points.
To prove part (2) correctly, we use the fact that $g(x) = x$ and apply the triangle inequality to obtain
$$\vert S(g\,;\dot{\mathcal{Q}}) - \frac{1}{2} \vert=\vert S(g\,;\dot{\mathcal{Q}}) - S(g\,;\dot{\mathcal{P}}) \vert= \left|\sum_{i=1}^n[g(q_i) - g(t_i)](x_i-x_{i-1})\right| \\ = \left|\sum_{i=1}^n[q_i-t_i](x_i-x_{i-1})\right| \leqslant \sum_{i=1}^n|q_i - t_i|(x_i-t_i)$$
Since it is given that $\| \dot{\mathcal{Q}}\| \lt \varepsilon$, we have $|q_i - t_i| \leqslant x_i - x_{i-1}\leqslant \| \dot{\mathcal{Q}}\| \lt \varepsilon$ for every index $i$, and, thus,
$$\vert S(g\,;\dot{\mathcal{Q}}) - \frac{1}{2} \vert \leqslant \sum_{i=1}^n|q_i - t_i|(x_i-t_i) \leqslant \varepsilon\sum_{i=1}^n (x_i - x_{i-1}) = \varepsilon$$
