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For finitely many Lebesgue measurable sets $A_i\subseteq [0,1]$ and nonnegative demands $b_i$, pairwise disjoint measurable subsets $B_i\subseteq A_i$ with $\lambda (B_i)=b_i$ exist exactly when $\lambda ({\bigcup }_{i\in I}{A}_i)\ge {\sum }_{i\in I}{b}_i$ for every index subset $I$. Necessity is additivity and monotonicity of Lebesgue measure. For sufficiency, partition the set union into membership cells $E_J$, send flow from a source through demand vertices $i$ to cells with $i\in J$, then to a sink. Source capacities are $b_i$, cell capacities are $\lambda (E_J)$, and intermediate capacities are $D={\sum }_i{b}_i$. Every cut of a flow network has capacity at least $D$ by the assumed inequalities. The max-flow min-cut theorem provides the allocations, and divisibility of Lebesgue measure turns each cell allocation into disjoint pieces. If $D=0$, all $B_i$ can simply be empty.

Write $D={\sum }_{i=1}^n{b}_i$ and let $\lambda$ denote Lebesgue measure. Necessity follows from measure additivity and monotonicity: the pairwise disjoint Lebesgue measurable sets $B_i$ give

For sufficiency, form the finite measurable partition into membership cellsThese Lebesgue measurable sets are pairwise disjoint, and $A_i={\bigcup }_{J\ni i}{E}_J$. Empty cells may be retained. Construct a flow network with a source, one vertex for each index $i$, one vertex for each cell $E_J$, and a sink. Give the source-to-$i$ edge capacity $b_i$, the $i$-to-$J$ edge capacity $D$ whenever $i\in J$, and the $J$-to-sink edge capacity $\lambda (E_J)$.

Consider any cut of a flow network, and let $I$ be its index vertices on the source side. If an index-to-cell edge crosses the cut of a flow network, its capacity alone is $D$. Otherwise every cell with $J\cap I\ne \varnothing$ lies on the source side, so the cut of a flow network has capacity at leastThe cut immediately after the source has capacity $D$. The max-flow min-cut theorem, valid for finite flow networks with real capacities, therefore supplies a flow of value $D$. Its cell allocations $f_{iJ}\ge 0$ satisfy

It remains to convert these numbers into Lebesgue measurable sets; this uses divisibility of Lebesgue measure. Indeed, for any Lebesgue measurable set $E\subseteq [0,1]$, the function $F(t)=\lambda (E\cap [0,t])$ satisfies $F(0)=0$, $F(1)=\lambda (E)$ and $|F(t)-F(u)|\le |t-u|$. Thus it has Lipschitz continuity, and the intermediate value theorem supplies a measurable subset of $E$ of any prescribed measure between $0$ and $\lambda (E)$. Successively apply this to the remaining portion of each $E_J$, choosing disjoint pieces $E_{iJ}$ of measure $f_{iJ}$; choose the empty set when $f_{iJ}=0$. ThenThis proves the measurable Hall theorem. The splitting step is essential: an arbitrary measure with an atom of a measure would not support the same conclusion. Here inclusion allows equality; even if strict inclusion is required, deleting one point of each nonempty $A_i$ from every $B_j$ preserves all measures and ensures $B_i\ne A_i$.