One of the main applications of the Lebesgue integral is to the study of various sorts of functions space. These are vector spaces whose elements are functions of various types. One of the most important examples of a function space is the space of measurable functions whose absolute values are p^{th} power integrable where p ≥ 1. These spaces, referred to as L^{p} spaces, are very useful in applications. In the chapter (Ω,S,μ) will be a measure space.
For each p > 1 define q by

Often one uses p^{′} instead of q in this context.
L^{p}
Proof: First here is a proof of Young’s inequality .
Proof: Consider the following picture:
From this picture, the sum of the area between the x axis and the curve added to the area between the t axis and the curve is at least as large as ab. Using beginning calculus, this is equivalent to the following inequality.

The above picture represents the situation which occurs when p > 2 because the graph of the function is concave up. If 2 ≥ p > 1 the graph would be concave down or a straight line. You should verify that the same argument holds in these cases just as well. In fact, the only thing which matters in the above inequality is that the function x = t^{p−1} be strictly increasing.
Note equality occurs when a^{p} = b^{q}.
Here is an alternate proof.
Lemma 22.1.4 For a,b ≥ 0,

and equality occurs when if and only if a^{p} = b^{q}.
Proof: If b = 0, the inequality is obvious. Fix b > 0 and consider

Then f^{′}

It follows f ≥ 0 and this yields the desired inequality.
Proof of Holder’s inequality: If either ∫ f^{p}dμ or ∫ g^{p}dμ equals ∞, the inequality 22.1 is obviously valid because ∞≥ anything. If either ∫ f^{p}dμ or ∫ g^{p}dμ equals 0, then f = 0 a.e. or that g = 0 a.e. and so in this case the left side of the inequality equals 0 and so the inequality is therefore true. Therefore assume both ∫ f^{p}dμ and ∫ g^{p}dμ are less than ∞ and not equal to 0. Let

and let

Hence,

This proves Holder’s inequality.
The following lemma will be needed.
Lemma 22.1.5 Suppose x,y ∈ ℂ. Then

Proof: The function f
Now as shown above,

which implies

and this proves the lemma.
Note that if y = ϕ
Proof: If p = 1, this is obvious because it is just the triangle inequality. Let p > 1. Without loss of generality, assume

and

Now

The following follows immediately from the above.
This shows that if f,g ∈ L^{p}, then f + g ∈ L^{p}. Also, it is clear that if a is a constant and f ∈ L^{p}, then af ∈ L^{p} because

Thus L^{p} is a vector space and
a.)
b.)
c.)
f →
Definition 22.1.8 Let f ∈ L^{p}

Then with this definition and using the convention that elements in L^{p} are considered to be the same if they differ only on a set of measure zero,
The following is an important definition.
Definition 22.1.9 A complete normed linear space is called a Banach^{1} space.
L^{p} is a Banach space. This is the next big theorem.
Theorem 22.1.10 The following hold for L^{p}(Ω)
b.) If {f_{n}} is a Cauchy sequence in L^{p}(Ω), then there exists f ∈ L^{p}
Proof: Let {f_{n}} be a Cauchy sequence in L^{p}(Ω). This means that for every ε > 0 there exists N such that if n,m ≥ N, then f_{n} − f_{m}_{p} < ε. Now select a subsequence as follows. Let n_{1} be such that f_{n} − f_{m}_{p} < 2^{−1} whenever n,m ≥ n_{1}. Let n_{2} be such that n_{2} > n_{1} and f_{n} − f_{m}_{p} < 2^{−2} whenever n,m ≥ n_{2}. If n_{1},

Then by the corollary to Minkowski’s inequality,

for all m. It follows that
 (22.3) 
for all m and so the monotone convergence theorem implies that the sum up to m in 22.3 can be replaced by a sum up to ∞. Thus,

which requires

Therefore, ∑ _{k=1}^{∞}g_{k+1}(x) converges for a.e. x because the functions have values in a complete space, ℂ, and this shows the partial sums form a Cauchy sequence. Now let x be such that this sum is finite. Then define

since ∑ _{k=1}^{m}g_{k+1}(x) = f_{nm+1}(x) − f_{n1}(x). Therefore there exists a set, E having measure zero such that

for all x


 (22.4) 
Therefore, f ∈ L^{p}(Ω) because

and lim_{k→∞}f_{nk} − f_{p} = 0. This proves b.).
This has shown f_{nk} converges to f in L^{p}
In working with the L^{p} spaces, the following inequality also known as Minkowski’s inequality is very useful. It is similar to the Minkowski inequality for sums. To see this, replace the integral, ∫ _{X} with a finite summation sign and you will see the usual Minkowski inequality or rather the version of it given in Corollary 22.1.7.
To prove this theorem first consider a special case of it in which technical considerations which shed no light on the proof are excluded. For the notion of product measure used here see Problem 7 on Page 1540.
Lemma 22.1.11 Let (X,S,μ) and (Y,ℱ,λ) be finite complete measure spaces and let f be μ × λ measurable and uniformly bounded. Then the following inequality is valid for p ≥ 1.
 (22.5) 
Proof: Since f is bounded and μ

Let

Note there is no problem in writing this for a.e. y because f is product measurable. Then by Fubini’s theorem,
 (22.6) 
Therefore, dividing both sides by the first factor in the above expression,
 (22.7) 
Note that 22.7 holds even if the first factor of 22.6 equals zero. ■
Now consider the case where f is not assumed to be bounded and where the measure spaces are σ finite.
Theorem 22.1.12 Let (X,S,μ) and (Y,ℱ,λ) be σfinite measure spaces and let f be product measurable. Then the following inequality is valid for p ≥ 1.
 (22.8) 
Proof: Since the two measure spaces are σ finite, this means 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
