 
[not yet completed]
Martingales
4.1 Conditional Expectation
page 224
At the end of the proof following
example 1.5, the final integrand
should be g(X) instead
of g(Y).

page 226
For the centered equation in the
first proof on this page, you can
think of the inequality as
saying that
the pnorm of E(XF) <=
the pnorm of X.

4.2 Martingales, Almost Sure Convergence
page 231
Thinking about the definition Durrett
gives for a martingale:
in part (iii), why not just
replace F_{n} by
sigma(X_{1},...,
X_{n}) ? This would be
just as useful (as far as I can tell),
and not change the object being defined.
In fact, Breiman uses this alternative
definition.

page 234
About halfway down the page, the function
H_{n} is incorrectly defined.
It should be 1_{{N<n}}.

page 235
It is intuitively helpful to notice that
a + (X_{m}a)^{+}
is just a euphemism for max(X_{m},a).

4.3 Examples
page 240
The equality of corollary (3.2) is
almost surely (not an exact equality).
If you examine the proof, you'll notice
that it only shows the probability of
the equality holding is 1.

page 244
Kakutani dichotomy
This introductory paragraph before
theorem (3.5) can be a little confusing.
About halfway through, we see that
"(3.3) implies that X_{n}
>X nua.s." If you look at (3.3), it
is not entirely clear how it says this. It is
actually more pertinent to use (3.4) to see
that X_{n} is a martingale,
notice that X_{n} is nonnegative,
and apply (2.11) to see that the sequence does
in fact have a nua.s. limit. Next Durrett
mentions "the convergence of the infinite
product," and it is not entirely clear that we have
seen any infinite products here. But we have:
X_{n} is actually the product
of q_{m} for m=1..n
(this is later officially stated as one of the
displayed equations in the proof of (3.5).)

page 245
Near the end of the proof of (3.5), it is stated
that P(X=infty)=0. But we don't have a
prob measure P! The measure should be nu instead.

page 246
In the statement of theorem (3.7),
it should be: almost surely,
(Z_{n}=0 for all
n sufficiently large). Notice
that how large n needs to be
may very well depend on omega. That
is, the proof does not show that there
is an N so that for all
n>N, Z_{n}=0
almost surely.

