Solution
= Solution
Given $\varepsilon>0$, choose a partition of $[a,d]$ whose upper-minus-lower sum is below $\varepsilon$, and refine it by inserting $b,c$. The contribution from subintervals lying in $[b,c]$ is nonnegative and no larger than the total difference. Restricting the refined partition to $[b,c]$ therefore gives upper-minus-lower sum below $\varepsilon$, so the criterion proves integrability there.