Problems and Solutions in - Mathematics - University of Hawaii

1.1 1991 November 21 1 REAL ANALYSIS
so this is what we will prove. The Radon-Nikodym theorem (A.12) says, if λ ≪ m are σ-finite positive measures
on a σ-algebra Σ, then there is a unique g ∈ L1 (dm) such that
λ(E) = g dm, ∀E ∈ Σ.
E
In the present case, µ ≪ µ + ν (since the measures are positive), so the theorem provides an f ∈ L1 (µ + ν) such
that
µ(E) = f d(µ + ν) ∀ E ∈ Σ,
E
which proves (5). ⊓⊔
9. 12 If f and g are integrable functions on (X, S, µ) and (Y, T , ν), respectively, and F (x, y) = f(x) g(y), show that
F is integrable on X × Y and
F d(µ × ν) = f dµ g dν.
Solution: 13 To show F (x, y) = f(x)g(y) is integrable, an important (but often overlooked) first step is to prove
that F (x, y) = f(x)g(y) is (S ⊗ T )-measurable. Define Ψ : X × Y → R × R by Ψ(x, y) = (f(x), g(y)), and let
Φ : R × R → R be the continuous function Φ(s, t) = st. Then,
F (x, y) = f(x)g(y) = (Φ ◦ Ψ)(x, y).
Theorem A.2 states that a continuous function of a measurable function is measurable. Therefore, if we can show
that Ψ(x, y) is an (S ⊗T )-measurable function from X ×Y into R×R, then it will follow that F (x, y) is (S ⊗T )measurable.
To show Ψ is measurable, let R be an open rectangle in R × R. Then R = A × B for some open sets
A and B in R, and
Ψ −1 (R) = Ψ −1 (A × B)
= {(x, y) : f(x) ∈ A, g(y) ∈ B}
= {(x, y) : f(x) ∈ A} ∩ {(x, y) : g(y) ∈ B}
= (f −1 (A) × Y ) ∩ (X × g −1 (B))
= f −1 (A) × g −1 (B).
Now, f −1 (A) ∈ S and g −1 (B) ∈ T , since f and g are S- and T -measurable, resp. Therefore, Ψ −1 (R) ∈ S ⊗ T ,
which proves the claim.
Now that we know F (x, y) = f(x)g(y) is (S ⊗T )-measurable, we can apply part (b) of the Fubini-Tonelli theorem
(A.13) to prove that F (x, y) = f(x)g(y) is integrable if one of the iterated integrals of |F (x, y)| is finite. Indeed,
|f(x)g(y)| dν(y) dµ(x) = |f(x)||g(y)| dν(y) dµ(x)
X Y
X
Y
= |f(x)| |g(y)| dν(y) dµ(x)
X
Y
= |f(x)| dµ(x) |g(y)| dν(y) < ∞,
12 See also: November ’97 (2), and others.
13 I’m not sure if the claim is true unless the measure spaces are σ-finite, so I’ll assume all measure spaces σ-finite.
In my opinion, the most useful version of the Fubini-Tonelli theorem is the one in Rudin [8], which assumes σ-finite measure spaces. There is a
version appearing in Royden [6] that does not require σ-finiteness. Instead it begins with the assumption that f is integrable. To me, the theorem
in Rudin is much easier to apply. All you need is a function that is measurable with respect to the product σ-algebra S ⊗ T , and from there, in a
single theorem, you get everything you need to answer any of the standard questions about integration with respect to a product measure.
8
X
Y