consider the surjection that maps A to 1, B-A to 2, and W - B to 3 and then appeal to P2.) 
There are now two cases to consider. First, assume that Pr(AIB) < 1 is rational, say mjn. Now consider the space Mn+l· Let A= {1, . .. ,m}, B = {1, ... ,n}, 
and Pr give each of 1, ... , n equal probability. We can consider the surjection g : Wn+l -+ W3 that maps 1, . . . ,m to 1, m+ 1, ... ,n to 2, and n+ 1 to 3. Thus, by P2 again, the result is true for M3 if we can show that it is true for an arbitrary measure Pr on Mn+I 
and Pr' E UpdMn+l (Pr, B). Note that the reason we have to introduce both M3 and Mn+l is that there is always a mapping from any other M for which the proposition is relevant into the former space, but not necessarily into the latter. 

Clearly Pr' does not give equal probability to each of the points 1, ... , n (for if it did, we would have Pr'(A) = mjn). In fact, the average probability of a point in A (according to Pr') is 1/n- t:jm, where E = Pr(AIB)- Pr'(A). There are two cases to consider: If Pr(AIB) � 1/2, then n - m ::; m. In this case, let C consist of the n- m elements of A with the lowest probability (according to Pr'). We must have Pr'(C) ::; (n- m)jn- (n- m)t:jm. Let h be a permutation on {1, .. . , n + 1} that switches the points in C 
with the points in B-A such that h(n + 1) = n + 1. 

Note that h*(Pr) = Pr (since Pr gives equal probability to all the points in B). Let Pr" = h*(Pr'). 

Note that Pr"(B- A) ::; (n- m)jn- ((n- m)jn)t:, 
so Pr"(A) � mjn + ((n- m)jn)t:. If Pr(AIB) < l/2, then n - m > m, and a similar argument works: This time, let C consist of the m elements of A with the lowest probability. In this case we get that Pr"(A) � 
mjn + (mjn)t:. Since h*(Pr') E UpdM(Pr,B) by P2, we are done. Note that, in either case, Pr" (A) � 
Pr(AIB) + min(Pr(AIB), 1-Pr(AIB))t:. 

Next suppose that Pr(AIB) is irrational. Choose r rational such that Pr(AIB) > r > Pr(AIB) -min(r, 1r)t:/2. By using P2 again, it suffices to prove the result for M4, with A = {1, 2}, B = {1, 2, 3}, and Pr({1}IB) = r. Let A' = {1}. Since Pr'(A') :::; 
Pr'(A) = Pr(AIB)-t:, it follows that Pr'(A') < r-t:/2. 

Since Pr(A'IB) is rational by construction, by the previous argument, there exists Pr" E UpdM (Pr, B) such that Pr"(A') � r + min(r, 1-r)t:/2 > Pr(AIB). Since A'� A, we have Pr"(A) > Pr(AIB), as desired. I 
Proposition 4.5: Suppose that Upd satisfies P1-
P5 and that UpdM(Pr,B) = 0 for some space M and Pr(B) \# 0. Then Upd = Updconstrain· 
We prove this using Lemmas A.1-A.3. 

Lemma A.l: If Upd satisfies P2, UpdM(Pr,B) = 0, and Pr(B) = a < l, then for every space M', probability measure Pr' E AM', qnd B' E F' such that Pr'(B') = a, we have UpdM' (Pr',B') = 0. 

Proof: First assume that M' = M2 B' = {2}. Then the result is immediate using P2. (Consider the M-M' representation shift that maps B to {2} and B to 
{1}.) Now if M' is arbitrary, we again get the result by applying P2 and using the fact that it holds for M2. 

(Again, consider the M'-M2 representation shift that maps B' to {2} and B to {1}.) I 
We can bootstrap our way up to an even stronger lemma. 

Lemma A.2 : If Upd satisfies P2 and P3, and UpdM(Pr,B) = 0 for some B with Pr(B) =a < 1, then for every domain M', probability measure Pr' E 
AM
', and B' � W' such that Pr' (B') < a, we have UpdM' (Pr', B') = 0. 

Proof: First consider the space M3 and let Pr" be a measure such that Pr" ( {2, 3}) = a and Pr"({3}) = {3 < a. By Lemma A.1, we have that UpdM3 (Pr", {2, 3}) = 0. By P3, we have that UpdM3(Pr",{2}) = UpdM3((UpdM3(Pr",{2,3}),{2})) 
= UpdM3(0, {2}) = 0. 

By Lemma A.1 again, it follows that for every space M' = (W', F'), probability measure Pr' E AM', 
and B' E F' such that Pr'(B') = {3, we have UpdM' (Pr', B') = 0. Since {3 was arbitrary, the desired result follows. I 
Finally, we get the strongest possible result of this type. 

Lemma A.3: If Upd satisfies P1, P2, P3, and PS, 
UpdM (Pr, B) = 0, then UpdM' (Pr', B') = 0 for every domain M', probability measure Pr' E AM', and set B' such that Pr' ( B') < 1. 

Proof: Let "Y* = supM
,,s{Pr'(B'): UpdM'(Pr',B) = 
0}. We want to show that "Y* = 1. Suppose by way of contradiction that "Y* < 1. Choose € > 0 such that 0 < "Y* - E, "Y* + E < 1, and ("'!* - t:)/("Y* + t:) > "Y*. 

Consider the space M3 again and let Pr' be such that Pr'(2) = "Y*- € and Pr'(3) = 2t:. By choice of "Y* and Lemma A.2, we have that UpdM3 (Pr', {2}) = 0 and UpdM3 (Pr', {2, 3}) \# 0. By P3, we have 

$$\begin{array}{l}{{U_{P p d}{}^{M_{3}}((\,U_{P p d}{}^{M_{3}}(\mathrm{Pr}^{\prime},\{2,3\}),\{2\})}}\\ {{=\,U_{P p d}{}^{M_{3}}(\mathrm{Pr}^{\prime},\{2\})=\varnothing.}}\end{array}$$
$$\left(2\right)$$

However, since UpdM3 (Pr', {2, 3}) \# 0, by Proposition 4.3, there exists some Pr" E UpdM3 (Pr', {2, 3}) 
such that Pr"(2) � Pr'( {2}1{2, 3}) = ("'!* - t:)/b* + 
.:) > "Y*· By Lemma A.2 and the choice of 
"Y\ we have that UpdM3(Pr",{2}) \# 0. By P5, UpdM3(Pr",{2}) � UpdM3((UpdM3(Pr',{2,3}),{2}). 

This contradicts (2). Thus, we must have "Y* = 1. I 