There is an extremely useful inequality. To prove this theorem first consider a special case of it in which
technical considerations which shed no light on the proof are excluded.

Lemma 7.5.1Let (X,S,μ) and (Y,ℱ,λ) be finite complete measure spaces and let f be μ×λ measurableand uniformly bounded. Then the following inequality is valid for p ≥ 1.

∫ (∫ ) 1p (∫ ∫ )1p
|f(x,y)|pdλ dμ ≥ ( |f (x,y)|dμ)pdλ . (7.13)
X Y Y X

(7.13)

Proof:Since f is bounded and μ

(X)

,λ

(Y )

< ∞,

( ∫ ∫ p ) 1p
Y( X |f(x,y)|dμ) dλ < ∞.

Let

∫
J(y) = |f(x,y)|dμ.
X

Note there is no problem in writing this for a.e. y because f is product measurable. Then by Fubini’s
theorem,

∫ ( ∫ )p ∫ ∫
|f(x,y)|dμ dλ = J(y)p− 1 |f(x,y)|dμdλ
Y X ∫Y ∫ X
p− 1
= X Y J(y) |f(x,y)|dλdμ

Now apply Holder’s inequality in the last integral above and recall p − 1 =

p
q

. This yields

∫ (∫ )p
|f (x,y)|dμ dλ
Y X
∫ (∫ )1q (∫ ) 1p
≤ J(y)pdλ |f(x,y)|pdλ dμ
X Y Y
(∫ p )1q ∫ (∫ p ) 1p
= Y J(y) dλ X Y |f(x,y)|dλ dμ

(∫ ∫ ) 1q ∫ ( ∫ ) 1p
= ( |f (x,y)|dμ)pdλ |f(x,y)|pdλ dμ. (7.14)
Y X X Y

(7.14)

Therefore, dividing both sides by the first factor in the above expression,

( ( ) ) 1 ( ) 1
∫ ∫ p p ∫ ∫ p p
Y X |f(x,y)|dμ dλ ≤ X Y |f(x,y)| dλ dμ. (7.15)

(7.15)

Note that 7.15 holds even if the first factor of 7.14 equals zero. ■

Now consider the case where f is not assumed to be bounded and where the measure spaces are σ
finite.

Theorem 7.5.2Let (X,S,μ) and (Y,ℱ,λ) be σ-finite measure spaces and let f be product measurable.Then the following inequality is valid for p ≥ 1.

∫ (∫ p ) 1p (∫ ∫ p )1p
X Y |f(x,y)| dλ dμ ≥ Y( X |f (x,y)|dμ) dλ . (7.16)

(7.16)

Proof: Since the two measure spaces are σ finite, there exist measurable sets, X_{m} and
Y_{k} such that X_{m}⊆ X_{m+1} for all m, Y_{k}⊆ Y_{k+1} for all k, and μ

(Xm )

,λ

(Yk )

< ∞. Now
define

{
fn(x,y) ≡ f (x,y) if |f (x,y)| ≤ n
n if |f (x,y)| > n.

Thus f_{n} is uniformly bounded and product measurable. By the above lemma,

∫ ( ∫ p )p1 ( ∫ ∫ p ) 1p
X Y |fn(x,y)| dλ dμ ≥ Y ( X |fn(x,y)|dμ) dλ . (7.17)
m k k m

(7.17)

Now observe that

|fn (x,y)|

increases in n and the pointwise limit is

|f (x,y)|

. Therefore, using the
monotone convergence theorem in 7.17 yields the same inequality with f replacing f_{n}. Next let k →∞ and
use the monotone convergence theorem again to replace Y_{k} with Y . Finally let m →∞ in what is left to
obtain 7.16. ■

Note that the proof of this theorem depends on two manipulations, the interchange of the order of
integration and Holder’s inequality. Note that there is nothing to check in the case of double sums. Thus if
a_{ij}≥ 0, it is always the case that