Finite Jordan covers bound upper integrals, while interior-disjoint Jordan subfamilies bound lower integrals

Extend all functions by zero to one bounding rectangle. The indicators in [L2] are integrable, and the nonnegativity of every $M_i$ makes the following inequality valid both on and off $E$. Monotonicity and finite linearity from [L1] give the cover bound, because each indicator integral is the content of its Jordan set: $h\le\sum_{i=1}^N M_i\mathbf1_{E_i}.$

The restriction of $h$ to each $F_i$ is integrable: away from grid cells meeting $\partial F_i$ its Darboux gap is inherited from $h$, while [L2] makes the total volume of boundary cells arbitrarily small. Hence $\int_{F_i}h\ge m_i\operatorname{cont}(F_i)$ by [L1]. Pairwise interior-disjoint Jordan sets intersect only on their content-zero boundaries, so the sum of their zero-extended restrictions equals the restriction to their union outside a content-zero set. The same boundary-cell argument and linearity [L1] therefore add these integrals without overcounting; their union lies in $E$, and $h\ge0$, giving the lower bound.

The boundary-cell argument in step 1.2 also makes the restriction of a signed integrable $h$ to $F$ integrable. A nonempty $F$ makes $\sup_F|h|$ a real number, since $h$ is bounded. On $F$, the inequalities $-|h|\le h\le|h|\le\sup_F|h|$ and [L1], together with the indicator identity in [L2], give the last estimate.