Problem 1
Part a
Suppose that f : X → C is measurable and g : Y → C is measurable, and let
h(x, y) = f (x)g(y). Consider a compact set E ⊂ C not containing zero (such
sets generate the σ-algebra of measurable sets in C). We need to show that
h−1 (E) is in the σ-algebra M ⊗ N . h−1 (E) is the set
{(x, y) : h(x, y) ∈ E}
which can be rewritten as
{(x, y) : f (x)g(y) ∈ E}
Or equivalently as
{(x, y) : f (x) =
p
for some p ∈ E. }
g(y)
(This is where we use the assumption that 0 ∈
/ E). This is the same as the
statement
{(x, y) : ∃a sequence zn of points in Q + iQ and a sequence pn of points in E
1
pn
1
such that |f (x) − zn | < and|
− zn | <
n
g(y)
n
for all n. This equivalence follows from the compactness of E- a subsequence of
p
. We can rewrite this
the pn ’s converges to a point p in E at which f (x) = g(y)
last set as
∞
\
[
{(x, y) : |f (x) − z| ≤
n=1 z∈Q+iQ
1
p
1
and |
− z| ≤ }
n
g(y)
n
where the sets inside are measurable by the definition of the product measure
and the measurability of f and g.
Part b
Note that f and g must have σ-finite support because they are in L1 . h is
supported on the product of the supports Sf of f and Sg of g, which is seen
to be σ-finite: we can write it as a countable union of the rectangles whose
sides are the sets approximating Sf and Sg . Furthermore, this set is measurable
because h is measurable. Therefore, we can view f and g as functions on the
σ-finite measure spaces (Sf , M|Sf , µ) and (Sg , N |Sg , ν), and we can view h as
a function on the product of these measure spaces. Fubini’s theorem therefore
applies to this product.
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Problem 2
Note that f (x, y) = (1 − xy)−a is a nonnegative function for any a. Therefore,
Tonelli’s theorem applies and these 3 integrals exist provided that any one of
1
them does. We can compute the integral in x first, giving (1−a)y
(1−(1−y)−a+1 )
1
if a 6= −1 and − y ln(1 − y) if a = 1. Note that this function is integrable when
−a + 1 > −1; therefore, we need a < 2.
Problem 3
Part a
Note that the integrand changes sign whenever x is a multiple of π. Furtherx|
on the interval [(n)π, (n + 1)π]. Note
more, the integrand is larger than | sin
nπ
that | sin x| is invariant under translations by multiples of π, so the integral
R (n+1)π
| sin(x)| is independent of n and nonzero (in fact, it is equal to 2);
(n)π
P∞ 2
therefore, we can bound the Riemann integral below by the sum
n=1 nπ ,
which diverges.
This proves that the Lebesgue integral is infinite by, e.g., continuity of measure applied to the intervals (0, n].
Part b
Notice that the value of the integral
Z ∞
e−xy sin(x) dy
0
is precisely
sin(x)
x ,
for any nonzero x. Consider the integral
Z
b
0
Z
∞
e−xy sin x dy dx.
0
We can apply Fubini’s theorem to this integral because it is bounded above by
the integral of sinx x from 0 to b, which is finite because sinx x has a finite limit as
x → 0. The above integral is therefore equal to
Z
∞
0
Z
b
e−xy sin x dx dy.
0
The inner integral can be evaluated using integration by parts; the antiderivative
−xy
−by
x+cos x)
b−e−by cos b
; plugging in b and 0 for x gives 1−e y sin
.
comes out to be − e (yysin
2 +1
y 2 +1
Now, we desperately hope that we can interchange the limit in b with the
outer integral in order to vastly simplify this expression. Note that for all b > 0
and all y > 0, we have the bound ye−by ≤ K for some constant K, so the
numerator is trapped between two constants. Therefore, for any b, the function
2
3
is bounded above in absolute value by 1+y
2 , which is integrable. So no matter
which sequence of b’s you pick, the dominated convergence theorem applies and
we can therefore interchange the limit and integral. This gives
Z ∞
1
dy
2
y +1
0
which is equal to
π
2
because the antiderivative is just arctan y.
Problem 4
Part a
First, I will show that ν + (E) is larger than the supremum of ν(F ) for all measurable F ⊂ E: to see this, perform a Hahn decomposition of E, and notice
that ν(F ) = ν(F ∩ P ) + ν(F ∩ N ) ≤ ν(F ∩ P ) ≤ ν(P ) = ν + (E), where the
last equality follows from the definition of ν + in the Jordan decomposition theorem. It’s easy to see that this supremum is attained and equal to ν + (E) just
by taking F = P . Taking F = N gives the result for ν − .
Part b
Clearly |ν|(E) is no larger than the supremum: just perform the decomposition
E1 = P
. Conversely, if E1 , . . . , Ej is aP
decomposition of E, we can
Pand E2 = NP
write j |ν(Ej )| = j |ν(Ej ∩ P ) + ν(Ej ∩ N )| ≤ j ν(Ej ∩ P ) − ν(Ej ∩ N ) =
P
P +
−
j |ν|(Ej ) = ν(E) as desired.
j ν (Ej ) + ν (Ej ) =
Problem 5
I’m assuming that µ1 , µ2 , ν1 , and ν2 are positive measures.
I will first show that ν1 × ν2 is absolutely continuous with respect to µ1 × µ2 .
Suppose that ν1 × ν2 (E) = M > 0 but µ1 × µ2 (E) = 0. Assume that E is
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contained in a rectangle on which both Radon-Nikodym derivatives dµ
dν1 and
dµ2
dν2 are bounded by some number B (If not, just discard the part of E where
this fails; if B is sufficiently large we are still left with a set of positive νmeasure because because the Radon-Nikodym derivative is finite a.e.). Then for
any countable collection
P of rectangles Rj covering E (arranged by decreasing
area), we have that j ν1 × ν2 (Rj ) ≥ M ; however, we can select collections
of rectangles (again contained in the product of the places
P where the RadonNikodym derivatives satisfy the uniform bound) so that j µ1 × µ2 (Rj ) < ǫ
for any ǫ > 0. Therefore, by the pigeonhole principle, at least one Rj , say,
(1)
(1)
µ1 ×µ2 (Rj∗ ). Therefore, either µ1 (Rj ∗ )/ν1 (Rj ∗ ) or
Rj ∗ satisfies ν1 ×ν2 (Rj ∗ ) ≥ M
ǫq
(2)
(2)
µ2 (Rj ∗ )/ν2 (Rj ∗ ) is at least M
ǫ . This can be made arbitrarily large by taking
ǫ sufficiently small, which contradicts the Lebesgue-Radon-Nikodym bound.
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It remains to be seen that the Radon-Nikodym derivative of µ1 × µ2 with
respect to ν1 × ν2 is the product
Radon-Nikodym
We clearly
R of 1the
R dµderivatives.
×µ2
dµ2
1
have for any rectangle that R dµ
(x
,
x
)
=
(x
)
(x
).
Because
1
2
1 dν2
2
dν1 ×ν2
R dν1
these rectangles generate the product σ-algebra of X1 × X2 , this implies that
the equality holds over general sets and not just rectangles. This implies that
the functions are equal almost everywhere by Proposition 2.23b in Folland.
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