I think this is OK (though still lots of room for subtleties). Spelling this aspect out in more detail:
Fix some arbitrary A which is strictly dominated by B.
We claim that there exists a face F and a continuous B’ over F such that B’ also dominates A.
Sample some lottery from B to obtain a concrete lottery b that strictly dominates A.
If b is a vertex we are done. Otherwise, let F be the face such that b lies in the interior of F.
For each voter, their level sets are either hyperplanes in F or else they are all of F.
We can ignore the voters who are indifferent within all of F, because any B’ supported in F will be the same as b from those voters’ perspectives.
Now define Xu,X′u,Lu,L as before, but restricting to the voters who have preferences within F.
We obtain a continuous distribution B’ for which f(B′,A)≈f(b,A) if we ignore the voters who were indifferent. But f(B′,A)=f(b,A) for the voters who are indifferent, so we have f(B′,A)≈f(b,A) overall.
(Of course this just goes through the existence of an open set of lotteries all of which strictly dominate A, we can just take B’ uniform over that set.)
This lemma is what we need, because we will run follow the leader over the space of pairs (F, A) where F is a face and A is a distribution over that face. So we conclude that the limit is not dominated by any pair (F, B’) where F is a face and B’ is a continuous distribution.
I still haven’t understood all of your argument, but have you missed the fact that some faces are entirely contained in L?
(Your arguments look similar to stuff we did when trying to apply this paper.)
I think this is OK (though still lots of room for subtleties). Spelling this aspect out in more detail:
Fix some arbitrary A which is strictly dominated by B.
We claim that there exists a face F and a continuous B’ over F such that B’ also dominates A.
Sample some lottery from B to obtain a concrete lottery b that strictly dominates A.
If b is a vertex we are done. Otherwise, let F be the face such that b lies in the interior of F.
For each voter, their level sets are either hyperplanes in F or else they are all of F.
We can ignore the voters who are indifferent within all of F, because any B’ supported in F will be the same as b from those voters’ perspectives.
Now define Xu,X′u,Lu,L as before, but restricting to the voters who have preferences within F.
We obtain a continuous distribution B’ for which f(B′,A)≈f(b,A) if we ignore the voters who were indifferent. But f(B′,A)=f(b,A) for the voters who are indifferent, so we have f(B′,A)≈f(b,A) overall.
(Of course this just goes through the existence of an open set of lotteries all of which strictly dominate A, we can just take B’ uniform over that set.)
This lemma is what we need, because we will run follow the leader over the space of pairs (F, A) where F is a face and A is a distribution over that face. So we conclude that the limit is not dominated by any pair (F, B’) where F is a face and B’ is a continuous distribution.
Ok, I believe this version of the Lemma, and am moving on to trying to get the rest of the argument.