Let $ f: [a,b] \rightarrow \mathbb{R} $ be bounded. And suppose for all $ a < x Theorem: Let $ f: [a,b] \rightarrow \mathbb{R} $ bounded. And suppose for all $ a < x <b $, $ f \in R([a,x]) $ . Then $ f \in R([a,b]) $.
Proof from lecture: Let $ \epsilon > 0 $. We'll show there exists $ \delta > 0 $ s.t. for every partition $ \prod $ of $ [ a, b] $ with $ \lambda ( \prod ) < \delta $ it occurs that $ \omega(f,\prod)< \epsilon $. We'll chose $ b - \frac{\epsilon}{4(M-m)} = b' < b $, where $ M,m $ are the bounds of $ 
f $ on $ [a,b] $ s.t. for every $ x \in [a,b], m \leq f(x) \leq M $.
Since $ f \in R([a,b']) $ there exists $ \delta_1 > 0 $ s.t. for every partition $ \prod_1 $  of $ [ a,b'] $ with $ \lambda(\prod_1)<\delta_1 $ it occurs that $ \omega(f|_{[a,b']},\prod_1) < \frac{\epsilon}{2} $ (here we just used Darboux's defintion for integrability ).
We'll chose $ \delta = min(\delta_1, \frac{\epsilon}{4(M-m)}) $ and let $\prod $ be arbitrary partition of $ [ a, b] $ with $ \lambda(\prod) < \delta $. Denote $ \prod_1 = ( \prod \cap[a,b'] )\cup \{ b' \} $ , this is a partition of $ [ a,b' ] $.
There exists $ n_0 $ s.t. $ x_{n_0} \leq b' $ .  Thus,
$ \omega(f,\prod) \leq \omega(f|_{[a,b']},\prod_1) + \omega(f,[b'-\delta,b]) \underbrace{\leq}_{ \lambda(\prod_1) < \delta \leq \delta_1} \frac{\epsilon}{2} + (M -m)(b-b'-\epsilon) \leq \epsilon $.  $ \square$
Note:  we could've wrote $ \omega(f,[x_{n_0},b]) $ instead of $  \omega(f,[b'-\delta,b]) $ but the latter  is a bigger interval so we prefer to take that one.
Definition reminders:
$ \lambda(\prod) = max_{i=1,...,n}|{ \triangle x_i}| $
$\omega(f, J)=\sup _{J} f-\inf _{J} f=\sup _{x, y \in J}(f(x)-f(y))$, where J is an interval.
$\omega(f, \Pi)=\sum_{i=1}^{n} \omega\left(f,\left[x_{i-1}, x_{i}\right]\right) \Delta x_{i}$ , where $ \Pi $ is a partition of some closed interval.
My question:
I don't understand the transition $ \omega(f,\prod) \leq \omega(f|_{[a,b']},\prod_1) + \omega(f,[b'-\delta,b]) $, can you please explain why the inequality holds? I'm having a little bit of difficulty wrapping my head around the arithmetic $ \omega(f|_{[a,b']},\prod_1) + \omega(f,[b'-\delta,b]) $.
 A: We have $\Pi = (x_0,x_1, \ldots x_n)$. By construction there exists an index $k< n$ such that $x_{k-1} < b' \leqslant x_k$. Since the partitions $\Pi$ and $\Pi_1$ have the points $x_0,x_1, \ldots, x_{k-1}$ in common it follows that
$$\omega(f, \Pi) - \omega(f, \Pi_1) \\= \omega(f,[x_{k-1},x_k])(x_k- x_{k-1}) - \omega (f, [b',x_k])(x_k-b') + \sum_{j= k+1}^n\omega(f, [x_{j-1},x_j])(x_j-x_{j-1}) $$
Note that for $j\geq k$, we have $\omega(f, [x_{j-1},x_j])\leqslant \sup_{x,y \in [b'-\delta,b]}|f(x) - f(y)|$ and, thus,
$$\omega(f,[x_{k-1},x_k])(x_k- x_{k-1}) - \omega (f, [b',x_k])(x_k-b')\sum_{j= k+1}^n\omega(f, [x_{j-1},x_j])(x_j-x_{j-1}) \\\omega(f,[x_{k-1},x_k])(x_k- x_{k-1})+  \sup_{x,y \in [b'-\delta,b]}|f(x) - f(y)| \cdot (b-x_k)\\ \leqslant  \sup_{x,y \in [b'-\delta,b]}|f(x) - f(y)|(b- x_{k-1}) = \omega(f,[b'-\delta,b])(b- x_{k-1})$$
We also have $b' - x_{k-1} \leqslant x_k - x_{k-1} \leqslant \delta$ and, hence, $b- x_{k-1} \leqslant b- (b'- \delta)$. This implies that
$$\omega(f, \Pi) - \omega(f, \Pi_1)\leqslant   \omega(f,[b'-\delta,b])(b- (b'-\delta)),$$
and, the correct inequality should be
$$\omega(f, \Pi) \leqslant  \omega(f, \Pi_1)+   \omega(f,[b'-\delta,b])[(b- b') + \delta]$$
Since $ \omega(f,[b'-\delta,b])\leqslant M-m$, $b - b' = \frac{\epsilon}{4(M-m)}$, and $\delta \leqslant  \frac{\epsilon}{4(M-m)}$, we get
$$\omega(f,[b'-\delta,b])[(b- b') + \delta] \leqslant (M-m)\left(\frac{\epsilon}{4(M-m)}+ \frac{\epsilon}{4(M-m)} \right) = \frac{\epsilon}{2},$$
and this facilitates completion of the proof.
