Suggestions for fixing up Wald material:

1. We could define the sample spaces for the various
experiments involving potentially infinite numbers of trials so that
they completely exclude the infinite trial case. 
That is, just consider the terminating executions in the probability
space.  
This seems reasonable because in the natural uncountable 
probability spaces that arise in the examples in 6042,
the set of infinite executions has probability 0 anyhow.
Furthermore, in the statement of Wald's theorem, the fact that E[Q] is
finite implies that the infinite executions have probability zero.

2. If we do this, then we can simplify Albert's Theorem 1.1 in his
handout 56 so that it doesn't mention the probability case.
Then it doesn't need to mention the exceptional case where V is
defined as 0.
We have to see if we can apply this version where we need it, however.
At the moment, the only place I see where we use this is in Wald's
theorem.
If Wald's Theorem is considered with the sample space defined to
exclude the nonterminating executions, then the simplified Theorem 1.1
seems to be sufficient for application within the proof.

3. But then we probably should restate Wald as follows:

``(Fix a countable sample space and outcome probabilities.)
Let Q be a random variable with values in N and with E[Q] finite.
Let R be a random variable with E[R] finite.
Let R1,R2,R3,...be a sequence of random variables such that:
  For each i, range(Ri) subseteq range(R)
  For each i and r, Pr(Ri = r | Q geq i) = Pr(R = r).
Let T = R1 + R2 + ...
Then E[T] = E[Q] E[R].''

This is simpler than what's there.

4. In order to prove the new version, we need to justify a step saying
that E[Ri Ii] = E[R] E[Ii].

It seems to me we can do this by brute force, as follows, without
using Tom's independence hypothesis:

E[Ri Ii]
= Sigma_{r in range(Ri), b in {0,1}}  r b Pr(Ri = r and Ii = b)
= Sigma_{r in range(Ri)}  r  Pr(Ri = r and Ii = 1)
    because all the b = 0 stuff drops out
= Sigma_{r in range(R)}  r  Pr(Ri = r and Ii = 1)

E[R] E[Ii] 
= Sigma_{r in range(R)}  r Pr(R = r) Pr(Ii = 1)

Thus, the needed equivalence reduces to showing:

For all i,r,
Pr(Ri = r and Ii = 1) = Pr(R = r) Pr(Ii = 1)

But 
Pr(Ri = r and Ii = 1) = Pr(Ri = r | Q geq i)  Pr(Q \geq i)
  by standard conditional probability,
= Pr(R = r) Pr(Q geq i) by the assumption in 3 above,
= Pr(R = r) Pr(Ii = 1), as needed.


5. Once we have shown this step, I think the rest of the proof is
straightforward.



