Here is a list corrections (grammtical, stylistic and
mathematical) in the first printing of "Fractals in
Probability and Analysis".
-----------------------------------------------------
Line 3 on page 33 of http://www.math.stonybrook.edu/~bishop/fractalbook_final.pdf
was-->were
Lemma 5.1.2 Add that $\alpha \leq 1$. (found by Max
Genkin)
page 143, line 11: remove reference to Exercise 5.19.
page 144, line -10: extra factor of $2$ needed in last
line of displayed equation, due to integral being over
two disjoint intervals.
page 145, line 5: The $C_1$ in the last term of the first
line of the displayed equation should have another name,
like M, and is an upper bound for $|f|$ on the interval
$[-\pi, \pi]$. Then $M$ should be included in the next two
lines and in line 9. The logic of the proof does not change.
(found by Moshe Stein).
page 150, line 11: use ``that book'' instead of ``this book''.
page 179, line 13: there is an uncecesary dot above ``One important ...''
page 207: line -1: ``proof of the upper boound'' should be
``proof of the lower bound''.
page 208: Corollary 7.2.2: we should assume $a>0$ in the second
part (clearly fails for $a=0$).
page 209: Wald's Lemma (Lemma 7.3.1) is incorrect as stated: one must assume
the stopping time $\tau$ has finite expectation in Part (i). Indeed,
a counterexample is discussed in the paragraph preceeding
the statement of the lemma. The correct version of part (i) is:
(i) $\expect(|B_\tau|^2) = \expect \tau$ if $\expect \tau < \infty$.
(ia) It suffice to prove this in the case $d =1$. The first part of the proof,
where we assume $\tau$ takes integer values and is bounded, is correct as
written in the book.
However, Parts (ib) and (ic) of the proof in the book should be replaced by
the following single argument.
(ib) If $\tau$ takes integer values (but need not be bounded), then
we know from part (a) that
$$\expect(|B_{\tau\wedge n}|^2) = \expect \tau \wedge n. $$
As $n \to \infty$ the right hand side tends to $\expect \tau $ (by
the Monotone Convergence Theorem). Also $B_{\tau\wedge n} \to
B_\tau$ almost surely (with respect to Wiener measure, since $\tau$
is finite almost surely). If $B_{\tau \wedge n}$ converges in $L^2$
(again, $L^2$ with respect to Wiener measure), then it must converge
in $L^2$ to $B_\tau$. Now
$$ \expect(B_{\tau\wedge(n+1)} | \field_n )
= B_{\tau\wedge n} + \expect[ B_{n+1} - B_n 1_{\tau \geq n}]
= B_{\tau\wedge n}, $$
so $B_{\tau\wedge m} - B_{\tau \wedge n} \perp B_{\tau
sedge n}$ for $m >n$. Thus
$$ \expect [ (B_{\tau \wedge m} - B_{\tau\wedge n)^2 ]
= \expect (B_{\tau \wedge m}^2 -\expect B_{\tau\wedge n^2
= \expect (\tau \wedge m} -{\tau\wedge n).$$
Hence $\{ B_{\tau\wedge n}$ is a Cauchy seqeunce in $L^2$.
Therefore $\{ \expect(|B_{\tau\wedge n}|^2) \}$ has a finite limit
and this must equal the limit of $\{\expect \tau \wedge n\} $$,
which we have seen is $\expect \tau$.
page 217, Figure 7.4.1: the first and third pictures on the
right side are the same; this is possible, but highly unlikely.
It seems that the same graphic accidentally got written into
two different file names, and we should insert a new pictures
into these positions.
page 232,line 1: $\Omega$ should be $f^{-1}(D)$.
page 242, Exercise 7.8 should be removed, since it is the same
as Corollary 7.9.3
page 316, Lemma 10.1.3: it was not clear to all readers why,
for segments, the integral has to be a multiple of the
length. Here is the argument in more detail.
Because the measure $\mu$ is invariant under rigid motions
of the plane the integral is the same for any two segments
of the same length. Thus it only depends on the length of the
segment. If we cut a unit segment into two equal segments,
one can be translated to the other, so they must have the same
integral, and by additivity these two equal values add to the
value of the integral over the whole segment; thus the integral
over each half segment is half the total integral. This works
whenever we divide the segment into n equal pieces, so a
segment k/n as long as the original has integral k/n as big;
i.e., we get a multiple of the identity on all segments whose
length is rational. Since the integral is increasing, we can
then deduce it is the identity for all values of the length.
page 317, Proof of Lemma 10.1.4: The rightmost inequality in
the dispplayed equation on line 5 of page 317 is poorly
explained. By the definition of Hausdorff measure, the
1-dimensonal measure of $E$ is the limit of $\sum r_j$
for some sequence of coverings of $E$ by disks. Hence we
can choose a covering of $E$ by disks so that the
corresponding radii sum is as close to the Hausdorff
$1$-measure as we wish, i.e., within a factor of 2.
Using this particular covering, we obtain the desired
inequality.
page 353, line 6: we mistakenly state that the analytic sets
form a sigma algegra. They do not, since the complement of an
analytic set need not be analytic. (found by Russ Lyons)
page 335, line 23: Carleson's $\epsilon^2$-conjecture is
now a theorem due to Benjamin Jaye, Xavier Tolsa
and Michele Villa in a 2019 preprint "A proof of Carleson's
$\epsilon^2$-conjecture, arXiv:1909.08581 [math.CA].