= Solution
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
$$
\lambda\!\left(\bigcup_{i\in I}A_i\right)\geq\lambda\!\left(\bigcup_{i\in I}B_i\right)=\sum_{i\in I}b_i.
$$
For sufficiency, form the finite <measurable partition> into membership cells
$$
E_J=\left(\bigcap_{j\in J}A_j\right)\setminus\left(\bigcup_{j\notin J}A_j\right),\qquad \varnothing\ne J\subseteq[n].
$$
These <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 least
$$
\sum_{i\notin I}b_i+\sum_{J:J\cap I\ne\varnothing}\lambda(E_J)=D-\sum_{i\in I}b_i+\lambda\!\left(\bigcup_{i\in I}A_i\right)\geq D.
$$
The 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 $f_{iJ}\geq0$ satisfy
$$
\sum_{J\ni i}f_{iJ}=b_i,\qquad \sum_{i\in J}f_{iJ}\leq\lambda(E_J).
$$
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 $|F(t)-F(u)|\leq|t-u|$. 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 $f_{iJ}=0$. Then
$$
\boxed{B_i=\bigcup_{J\ni i}E_{iJ}\subseteq A_i,\qquad \lambda(B_i)=b_i,\qquad B_i\cap B_j=\varnothing\ (i\ne j).}
$$
This 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 $B_i\ne A_i$.
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