11.8 Lebesgue Measure And Iterated Integrals
The following is the main result.
Theorem 11.8.1 Let f ≥ 0 and suppose f is a Lebesgue measurable function defined on
ℝn and ∫
ℝnfdmn < ∞. Then
This will be accomplished by Fubini’s theorem, Theorem ?? and the following
Lemma 11.8.2 mk × mn−k = mn on the mn measurable sets.
Proof: First of all, let R = ∏
i=1n(ai,bi] be a measurable rectangle and
let Rk = ∏
i=1k(ai,bi],Rn−k = ∏
i=k+1n(ai,bi]. Then by Fubini’s theorem,
and so mk × mn−k
agree on every half open rectangle. By Lemma 11.1.2
these two measures agree on every open set. Now if K
is a compact set, then
is the open set, K
Another way of saying this is
which is obviously open because
continuous function. Since
is the countable intersection of these decreasing open sets,
each of which has finite measure with respect to either of the two measures, it
follows that mk × mn−k
agree on all the compact sets. Now let E
bounded Lebesgue measurable set. Then there are sets, H
is a countable union of compact sets, G
a countable intersection of open
sets, H ⊆ E ⊆ G,
Then from what was just shown about
compact and open sets, the two measures agree on G
and on H.
By completeness of the measure space for mk × mn−k,
it follows E
is mk × mn−k
This proves the lemma.
You could also show that the two σ algebras are the same. However, this is not needed
for the lemma or the theorem.
Proof of Theorem 11.8.1: By the lemma and Fubini’s theorem, Theorem
Corollary 11.8.3 Let f be a nonnegative real valued measurable function.
Proof: Let Sp ≡
ℝnfXSpdmn < ∞.
Therefore, from Theorem 11.8.1
Now let p →∞ and use the Monotone convergence theorem and the Fubini Theorem ??
on Page ??.
Not surprisingly, the following corollary follows from this.
Corollary 11.8.4 Let f ∈ L1
where the measure is mn. Then
Proof: Apply Corollary 11.8.3 to the postive and negative parts of the real and
imaginary parts of f.