Sufficiency of Lebesgue's Criterion for Riemann Integrability

1. I think this is a trick that is used in this proof ( just like $\epsilon 2^{-n}$ trick used in a lot of measure-theoretic results).

2. Suppose $x_{j-1} =a$. Clearly $a \notin L$ so $x_j$ must be either $k_s+\delta_s$ or $k_s-\delta_s $ for some $1\leq s\leq r$ (Because the endpoints of $I_js$ are formed by the points in $P$). In either case, there exists $k_{s'} \in K$ such that $|I_j| \leq |(a+\delta_{s'})-a|$, where $s'\leq r$. By the choice of $\delta_{s'} (=\delta(k_{s'}))\,$, $|I_j| \leq \epsilon$.

3. $I_js $ have disjoint interiors. So the result follows from the countable-additivity, sub-additivity & the monotonicity of the Lebesgue measure.

$|\cup I_j| = \sum |I_j| \leq |\cup (a_j , b_j)| \leq \sum (b_j - a_j) < \epsilon$ (the last inequlity is due to the choice of $(a_j ,b_j)$s)