In the case that Upd satisfies P1-P5, P6', and P7, it follows from Proposition 4.5 that we must have I UpdM (Pr, B)l = 1 if Pr(B) f. 0. The fact that Upd = Updcond now follows immediately from Proposition 4.4. Thus, it remains to show that if if Upd satisfies P1-P5, P6", and P7, then Upd = Updcond
· 
To show this, we first prove an easy lemma, which shows that Upd must agree with conditioning at least on events with extreme probabilities. 

Lemma 4.6: If Upd satisfies P1, P3, and P4, Pr E 
D.M, A� B, and Pr(AIB) = 1 (resp., Pr(AIB) = 0), 
then for all Pr' E UpdM (Pr, B), we have Pr' (A) = 1 
(resp. , Pr'(A) = 0). 

Proof: Let M = (W, F). It suffices to prove the result for Pr(AIB) = 1, because the other case follows by considering B - A. 

Let C = W- (B - A). Since Pr(AIB) = 1, we have Pr( C) = 1. By P4, we have {Pr} = UpdM (Pr, C), 
and thus UpdM (Pr, B) = UpdM ( UpdM (Pr, C), B). 

By P3, UpdM( UpdM(Pr,C),B) = UpdM(Pr,B n C) = UpdM(Pr, A). It follows that UpdM(Pr, B) = 
UpdM(Pr,A). But by P1, if Pr' E UpdM(Pr, A), then Pr'(A) = 1. The result follows. I 
We now introduce a key concept. Given Upd, a measure space M = (W, F), a measure Pr E D.M, and events A C B E F such that 0 < Pr(A) :::; Pr(B) 
define 

$$U^{U_{p d,\mathcal{M},\Pr}}(A,B)=\operatorname*{sup}_{\Pr^{\prime}\in U_{p d}{}^{A\mathcal{M}}(\Pr,B)}{\frac{\Pr^{\prime}(A)}{\Pr(A)/\Pr(B)}}.$$

(We take U
Upd,M,
Pr(A, B) = -oo if UpdM(Pr,B) = 0.) 
The intuition is that U
Upd,M,
Pr(A, B) should be viewed as a measure of how far Upd is from Updcond
· It is the sup, over all the measures in UpdM (Pr, B), of the ratio of the (post-update) probability of A to Pr(AIB). 

Of course, if Upd = Updcond
> then uUpd,M,
Pr(A, B) = 
1 for all A, B. As the following result shows, the converse holds as well. 

Proposition 4. 7 : 
U
Upd,M,Pr(A, B) 
Upd = Updcond· 
If Upd satisfies P1-P4 and 1 for all M, A, and B, then Proof: Suppose that Upd f. Updcond
· Then there exists a measure space M = (W, F), Pr E D.M, 
B E F, A � B, and Pr' E UpdM(Pr,B) such that Pr'(A) f. Pr(AIB). By Lemma 4.6 we know that PrJ AlB) is neither 0 nor 1. If Pr' (A) > Pr(AIB), then U pd,M,
Pr(A, B) > 1, contradicting our assumption. 

But otherwise, we have Pr'(A) < Pr(AIB) and so, by Proposition 4.3, there exists Pr" E UpdM (Pr, B) such that Pr" (A) > Pr(AIB). Again, this contradicts our assumption. I 
In light of Proposition 4.7, we can complete the proof of Theorem 4.1 by proving the following result: 
Proposition 4.8: If Upd satisfies P1-P5, P6'', and P7, then U
Up d,M,
Pr(A, B) = 1 for all M, A, B. 

Despite the strength of P6", the proof of Proposition 4.8 turns out to be surprisingly difficult. (More accurately, we have not been able to find a proof that is not difficult!) The details are in the appendix. 

## 5 Discussion

The main purpose of this paper is to illustrate how different the set-of-measures model can be, technically, from the standard single-measure model. In general, it is unwise to simply assume that a result in the standard model can be trivially "lifted" to apply in general. 

In terms of understanding update procedures, one obvious next step would be to examine some of the other justifications for conditioning in the single-measure model, to see to what extent they carry over to the new setting. There are also outstanding questions even in the axiomatic framework considered here. P6, in particular, is quite a strong assumption. Is it really necessary? We conjecture that our main result actually holds with P6 replaced by P6*, although we have not been able to prove this. (Recall that P6' implies both P6" and P6* in the presence of P1 and P2.) 
We are clearly not advocating P6 (or even P6*) as being anywhere near as compelling as, say, P1 or P2. (Of course, one can raise reasonable arguments against P1 and P2-and most of the other postulates-as well.) 
So what does this say about Updcand and Updcanstrain? 

It seems quite plausible to us that other update procedures, such as UpdML• that will be appropriate in some circumstances. We believe that a more careful investigation into such alternative rules, and a continued effort to try and clearly determine what makes an update rule appropriate to a given domain, would be worthwhile. 

## A Appendix: Proofs

For the proofs, it is useful to define the family of spaces Mn = ( {1, ... , n }, 2{l, ... ,n}). 

Proposition 4.3: Suppose that Upd satisfies P1 and P2 and that for some Pr' E UpdM (Pr, B) and A such that 0 < Pr(AIB) < 1, we have Pr'(A) < Pr(AIB). 

Then there also exists some Pr" E UpdM (Pr, B) such that Pr"(A) > Pr(AIB). 

Proof: Consider M = (W, F) and assume that B f. 

W. (The argument if B = W is almost identical and left to the reader.) By P2, it suffices to prove the result in the case that M = M3. (For any other M, we can 