= Continuous-offer bargaining acceptance lemma
Let a proposer divide a pie of size $M>0$ by continuously offering its responder $q\in[0,M]$. Rejection gives fixed continuation <payoffs> $b$ to the proposer and $c$ to the responder, with $0\leq c<M$ and $b<M-c$. Then a <subgame perfect equilibrium> must have the responder accept $q=c$ with <probability> one. The proposer offers exactly $c$, retaining $M-c$.
Sequential rationality forces rejection for $q<c$ and acceptance for $q>c$. If acceptance at $c$ had probability $a<1$, that offer would give the proposer $a(M-c)+(1-a)b<M-c$. Every larger offer gives less than $M-c$ and can be improved by a smaller offer still above $c$; every smaller offer gives $b<M-c$. The <supremum> $M-c$ would therefore be unattained. Randomizing offers cannot attain it either: the nonnegative gap from $M-c$ would be strictly positive almost surely and hence have positive <expectation>. No <best response> would exist, contradicting equilibrium. Acceptance at $c$ removes this gap and makes $q=c$ the unique optimal proposal. The assumptions $c<M$ and $b<M-c$ exclude degenerate indifference of the proposer.
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