Let be bounded with integrable. If the set

is finite, then is integrable and .

Let and fix . Choose a partition of with

The idea is to isolate the problematic points with intervals of tiny total weight. Pick intervals with

and set and . Refining only improves , so .

Splitting the sum over subintervals inside and outside : if then there, so

if then merely is bounded, and

Thus , proving integrability of . For the values,

so ; the reverse inequality is identical, and since is arbitrary the integrals agree.