Proof. Recall that we know from Lem. 2 that

⎪ ⎪

ab

⎪ ⎪

≤

⎪ ⎪ a  ⎪ ⎪ p   p

+

⎪ ⎪ b  ⎪ ⎪ q   q

(a,b∈K).

Now we follow the steps in proof of Prop. 4. Define measurable functions a,b:X→K by setting

a(x) =

f(x) ⎪⎪ ⎪⎪ f ⎪⎪ ⎪⎪ p

,   b(x) =

g(x) ⎪⎪ ⎪⎪ g ⎪⎪ ⎪⎪ q

(x∈ X).

So we have that

⎪ ⎪

a(x) b(x)

⎪ ⎪

≤

⎪ ⎪ f(x)  ⎪ ⎪ p p  ⎪⎪ ⎪⎪ f ⎪⎪ ⎪⎪ pp

+

⎪ ⎪ g(x)  ⎪ ⎪ q q  ⎪⎪ ⎪⎪ g ⎪⎪ ⎪⎪ qq

(x∈ X).

By integrating, we see that

∫

X

⎪ ⎪

ab

⎪ ⎪

d µ ≤

1 p  ⎪⎪ ⎪⎪ f ⎪⎪ ⎪⎪ pp

∫

X

⎪ ⎪

f

⎪ ⎪

p  d µ +

1 q  ⎪⎪ ⎪⎪ g ⎪⎪ ⎪⎪ qq

∫

X

⎪ ⎪

g

⎪ ⎪

q  d µ =

1 p

+

1 q

= 1.

Hence, by the definition of a and b,

∫

X

⎪ ⎪

fg

⎪ ⎪

≤

⎪⎪ ⎪⎪

f

⎪⎪ ⎪⎪

p

⎪⎪ ⎪⎪

g

⎪⎪ ⎪⎪

q,

as required. □

Proof. Part 1 is easy. For 2, we need a version of Minkowski’s Inequality, which will follow from the previous lemma. We essentially repeat the proof of Prop. 5.

Notice that the p=1 case is easy, so suppose that 1<p<∞. We have that

∫   X   ⎪ ⎪ f+g  ⎪ ⎪ p  d µ =  ∫   X   ⎪ ⎪ f+g  ⎪ ⎪ p−1  ⎪ ⎪ f+g  ⎪ ⎪   d µ                      ≤  ∫   X   ⎪ ⎪ f+g  ⎪ ⎪ p−1  ⎛ ⎝ ⎪ ⎪ f  ⎪ ⎪  +  ⎪ ⎪ g  ⎪ ⎪   ⎞ ⎠  d µ                      =  ∫   X   ⎪ ⎪ f+g  ⎪ ⎪ p−1  ⎪ ⎪ f  ⎪ ⎪   d µ +  ∫   X   ⎪ ⎪ f+g  ⎪ ⎪ p−1  ⎪ ⎪ g  ⎪ ⎪   d µ.

Applying the lemma, this is

≤

⎪⎪ ⎪⎪

f

⎪⎪ ⎪⎪

p

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

∫

X

⎪ ⎪

f+g

⎪ ⎪

q(p−1)  d µ

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

1/q

+

⎪⎪ ⎪⎪

g

⎪⎪ ⎪⎪

p

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

∫

X

⎪ ⎪

f+g

⎪ ⎪

q(p−1)  d µ

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

1/q

.

As q(p−1)=p, we see that

⎪⎪ ⎪⎪

f+g

⎪⎪ ⎪⎪

pp ≤

⎛ ⎝

⎪⎪ ⎪⎪

f

⎪⎪ ⎪⎪

p +

⎪⎪ ⎪⎪

g

⎪⎪ ⎪⎪

p

⎞ ⎠

⎪⎪ ⎪⎪

f+g

⎪⎪ ⎪⎪

pp/q.

As p−p/q = 1, we conclude that

⎪⎪ ⎪⎪

f+g

⎪⎪ ⎪⎪

p ≤

⎪⎪ ⎪⎪

f

⎪⎪ ⎪⎪

p +

⎪⎪ ⎪⎪

g

⎪⎪ ⎪⎪

p,

as required.

In particular, if f,g∈ L_p(µ) then af+g∈ L_p(µ), showing that L_p(µ) is a vector space. □

Proof. We need to show that addition, and scalar multiplication, are well-defined on L_p(µ)/∼. Let a∈K and f₁,f₂,g₁,g₂∈ L_p(µ) with f₁∼ f₂ and g₁∼ g₂. Then it’s easy to see that af₁+g₁ ∼ af₂+g₂; but this is all that’s required!

If f ∼ g then | f |^p = | g |^p almost everywhere, and so ||f||_p = ||g||_p. So ||·||_p is well-defined on equivalence classes. In particular, if f∼ 0 then ||f||_p=0. Conversely, if ||f||_p=0 then ∫_X | f |^p d µ=0, so as | f |^p is a positive function, we must have that | f |^p=0 almost everywhere. Hence f=0 almost everywhere, so f∼ 0. That is,

⎧ ⎨ ⎩

f∈ Lp(µ) : f∼ 0

⎫ ⎬ ⎭

=

⎧ ⎨ ⎩

f∈ Lp(µ) :

⎪⎪ ⎪⎪

f

⎪⎪ ⎪⎪

p=0

⎫ ⎬ ⎭

.

It follows from the above lemma that this is a subspace of L_p(µ).

The above lemma now immediately shows that ||·||_p is a norm on L_p(µ)/∼. □

Proof. Consider first the case of a finite measure space X. We again follow the three steps scheme from Rem. 7. Let f_n be a Cauchy sequence in L_p(µ). From the Hölder inequality (80) we see that ||f_n−f_m||₁≤ ||f_n−f_m||_p (µ(X))^1/q. Thus, f_n is also a Cauchy sequence in L₁(µ). Thus by the Theorem 42 there is the limit function f∈ L₁(µ). Moreover, from the proof of that theorem we know that there is a subsequence f_nk of f_n convergent to f almost everywhere. Thus in the Cauchy sequence inequality

∫

X

⎪ ⎪

fnk −fnm

⎪ ⎪

p d µ <ε

we can pass to the limit m→ ∞ by the Fatou Lemma 39 and conclude:

∫

X

⎪ ⎪

fnk −f

⎪ ⎪

p d µ <ε.

So, f_nk converges to f in L_p(µ), then f_n converges to f in L_p(µ) as well.

For a σ-finite measure µ we represent X=⊔_k X_k with µ(X_k)<+∞ for all k. The restriction (f_n^(k)) of a Cauchy sequence (f_n)⊂L_p(X,µ) to every X_k is a Cauchy sequence in L_p(X_k,µ). By the previous paragraph there is the limit f^(k)∈ L_p(X_k,µ). Define a function f∈L_p(X,µ) by the identities f(x)=f^(k) if x∈ X_k. By the additivity of integral, the Cauchy condition on (f_n) can be written as:

∫

X

⎪ ⎪

fn−fm

⎪ ⎪

p d µ=

∞ ∑ k=1

∫

Xk

⎪ ⎪

fn(k)−fm(k)

⎪ ⎪

p d µ<ε.

It implies for any M:

M ∑ k=1

∫

Xk

⎪ ⎪

fn(k)−fm(k)

⎪ ⎪

p d µ<ε.

In the last inequality we can pass to the limit m→ ∞:

M ∑ k=1

∫

Xk

⎪ ⎪

fn(k)−f(k)

⎪ ⎪

p d µ<ε.

Since the last inequality is independent of M we conclude:

∫

X

⎪ ⎪

fn−f

⎪ ⎪

p d µ=

∞ ∑ k=1

∫

Xk

⎪ ⎪

fn(k)−f(k)

⎪ ⎪

p d µ<ε.

Thus we conclude that f_n→ f in L_p(X,µ). □

Proof. This proof very similar to proof of Thm. 13. For f∈ L_q(µ) and g∈ L_p(µ), it follows by the Hölder’s Inequality (80), that fg is summable, and

⎪ ⎪ ⎪ ⎪ ⎪ ⎪

∫

X

fg  d µ

⎪ ⎪ ⎪ ⎪ ⎪ ⎪

≤

∫

X

⎪ ⎪

fg

⎪ ⎪

d µ ≤

⎪⎪ ⎪⎪

f

⎪⎪ ⎪⎪

q

⎪⎪ ⎪⎪

g

⎪⎪ ⎪⎪

p.

Let f₁,f₂∈ L_q(µ) and g₁,g₂∈ L_p(µ) with f₁∼ f₂ and g₁∼ g₂. Then f₁g₁ = f₂g₁ almost everywhere and f₂g₁ = f₂g₂ almost everywhere, so f₁g₁ = f₂g₂ almost everywhere, and hence

∫

X

f1g1  d µ =

∫

X

f2g2  d µ.

So Φ is well-defined.

Clearly Φ is linear, and we have shown that ||Φ(f)|| ≤ ||f||_q.

Let f∈ L_q(µ) and define g:X→K by

g(x) =

⎧ ⎪ ⎨ ⎪ ⎩  f(x)    ⎪ ⎪ f(x)  ⎪ ⎪ q−2 : f(x)≠0,        0 : f(x)=0.

Then | g(x) | = | f(x) |^q−1 for all x∈ X, and so

∫

X

⎪ ⎪

g

⎪ ⎪

p  d µ =

∫

X

⎪ ⎪

f

⎪ ⎪

p(q−1)  d µ =

∫

X

⎪ ⎪

f

⎪ ⎪

q  d µ,

so ||g||_p = ||f||_q^q/p, and so, in particular, g∈L_p(µ). Let F=Φ(f), so that

F(g) =

∫

X

fg  d µ =

∫

X

⎪ ⎪

f

⎪ ⎪

q  d µ =

⎪⎪ ⎪⎪

f

⎪⎪ ⎪⎪

qq.

Thus ||F|| ≥ ||f||_q^q / ||g||_p = ||f||_q. So we conclude that ||F|| = ||f||_q, showing that Φ is an isometry. □

Proof.[Sketch of the proof] As µ(X)<∞, for E∈L, we have that ||χ_E||_p = µ(E)^1/p < ∞. So χ_E∈L_p(µ), and hence we can define

ν(E) = F(χE)    (E∈L).

We proceed to show that ν is a signed (or complex) measure. Then we can apply the Radon-Nikodym Theorem 53 to find a function f:X→K such that

F(χE) = ν(E) =

∫

E

f  d µ    (E∈L).

There is then a long argument to show that f∈ L_q(µ), which we skip here. Finally, we need to show that

∫

X

fg  d µ = F(g)

for all g∈ L_p(µ), and not just for g=χ_E. That follows for simple functions with a finite set of values by linearity of the Lebesgue integral and F. Then, it can be extended by continuity to the entire space L_p(µ) in view in the following Prop. 14. □

Proof.Let f∈L_p(µ), and suppose for now that f≥0. For each n∈ℕ, let

fn = min(n,

1 n

⌊ n f ⌋).

Then each f_n is simple, f_n ↑ f, and | f_n−f |^p→0 pointwise. For each n, we have that

0 ≤ fn ≤ f  0 ≤ f−fn ≤ f,

so that | f−f_n |^p ≤ | f |^p for all n. As ∫| f |^p  d µ<∞, we can apply the Dominated Convergence Theorem to see that

lim n

∫

X

⎪ ⎪

fn−f

⎪ ⎪

p  d µ = 0,

that is, ||f_n−f||_p → 0.

The general case follows by taking positive and negative parts, and if K=ℂ, by taking real and imaginary parts first. □

Proof. Let f ∈ L_p(ℝ,µ), since ∫_ℝ | f |^p d µ = ∑_k=−∞^∞ ∫_[k,k+1) | f |^p d µ there exists N such that ∑_k=−∞^−N + ∑_N^k=∞ ∫_[k,k+1) | f |^p d µ < ε . By the previous Proposition, the restriction of f to [−N,N] can be ε-approximated by a simple bounded function f₁ with support in [−N,N] attained only a finite number of values. Therefore f₁ will be also (2ε)-approximation to f as well. □

Proof. By Prop. 14, for a given f∈L_p(ℝ) and ε>0 there exists a simple function f₀=∑_k=1ⁿ c_k χ_Ak such that ||f−f₀||_p<ε/2. Let M=||f₀||_∞ < ∞. By measurability of the set A_k there is C_k=⊔_j^m_k [a_jk,b_jk) a disjoint finite union of half-open intervals such that µ(C_k▵ A_k)<ε/2n³ M. Since A_k and A_j are disjoint for k≠ j we also obtain by the triangle inequality: µ(C_j ∩ A_k)<ε/2n³ M and µ(C_j ∩ C_k)<2ε/2n³ M. We define a step function

f1=

n ∑ k=1

ck χCk=

n ∑ k=1

mk ∑ j

ck χ[ajk,bjk).

Clearly

f1(x)=ck    for all  x ∈ Ak∖ ((Ck▵ Ak) ⋃(⋃j≠ k Cj)).

Thus:

µ({x ∈ ℝ  ∣  f0(x)≠ f1(x)}) ≤ n· n·

ε 2n3 M

=

ε 2 n M

.

Then ||f₀−f₁||_p≤ nM· ε/2 n M=ε/2 because ||f₁||_∞< nM. Thus ||f−f₁||_p<ε. □

Proof. In view of Rem. 29 and the previous Lemma it is enough to show that the characteristic function of an interval [a,b] can be approximated by a continuous function in L_p(ℝ). The idea of such approximation is illustrated by Fig. 4 and we skip the technical details. □

Proof. As [0,1] is a finite measure space, and each member of C_K([0,1]) is bounded, it is easy to see that each f∈ C_K([0,1]) is such that ||f||_p<∞. So it makes sense to regard C_K([0,1]) as a subspace of L_p(µ). If C_K([0,1]) is not dense in L_p(µ), then we can find a non-zero F∈L_p([0,1])^* with F(f)=0 for each f∈ C_K([0,1]). This was a corollary of the Hahn-Banach theorem 15.

So there exists a non-zero g∈ L_q([0,1]) with

∫

[0,1]

fg  d µ = 0    (f∈ CK([0,1])).

Let a<b in [0,1]. By approximating χ_(a,b) by a continuous function, we can show that ∫_(a,b) g  d µ = ∫ g χ_(a,b)  d µ = 0.

Suppose for now that K=ℝ. Let A = { x∈[0,1] : g(x)≥0 } ∈ L. By the definition of the Lebesgue (outer) measure, for є>0, there exist sequences (a_n) and (b_n) with A ⊆ ∪_n (a_n,b_n), and ∑_n (b_n−a_n) ≤ µ(A) + є.

For each N, consider ∪_n=1^N (a_n,b_n). If some (a_i,b_i) overlaps (a_j,b_j), then we could just consider the larger interval (min(a_i,a_j), max(b_i,b_j)). Formally by an induction argument, we see that we can write ∪_n=1^N (a_n,b_n) as a finite union of some disjoint open intervals, which we abusing notations still denote by (a_n,b_n). By linearity, it hence follows that for N∈ℕ, if we set B_N = ⊔_n=1^N (a_n,b_n), then

∫

g χBN  d µ =

∫

g χ(a1,b1)⊔⋯⊔(aN,bN)  d µ = 0.

Let B=∪_n (a_n,b_n), so A⊆ B and µ(B) ≤ ∑_n (b_n−a_n) ≤ µ(A)+є. We then have that

⎪ ⎪

∫

g χBN  d µ −

∫

g χB  d µ

⎪ ⎪

=

⎪ ⎪

∫

g χB∖ (a1,b1)⊔⋯⊔(aN,bN)  d µ

⎪ ⎪

.

We now apply Hölder’s inequality to get

⎛ ⎝ ∫ χB∖ (a1,b1)⋃⋯⋃(aN,bN)  d µ  ⎞ ⎠ 1/p ⎪⎪ ⎪⎪ g ⎪⎪ ⎪⎪ q = µ(B∖ (a1,b1)⊔⋯⊔(aN,bN))1/p  ⎪⎪ ⎪⎪ g ⎪⎪ ⎪⎪ q                      ≤  ⎛ ⎜ ⎜ ⎝ ∞ ∑ n=N+1 (bn−an)  ⎞ ⎟ ⎟ ⎠ 1/p     ⎪⎪ ⎪⎪ g ⎪⎪ ⎪⎪ q.

We can make this arbitrarily small by making N large. Hence we conclude that

∫

g χB  d µ=0.

Then we apply Hölder’s inequality again to see that

⎪ ⎪

∫

gχA  d µ

⎪ ⎪

=

⎪ ⎪

∫

gχA  d µ −

∫

gχB  d µ

⎪ ⎪

=

⎪ ⎪

∫

g χB∖ A  d µ

⎪ ⎪

≤

⎪⎪ ⎪⎪

g

⎪⎪ ⎪⎪

q µ(B∖ A)1/p ≤

⎪⎪ ⎪⎪

g

⎪⎪ ⎪⎪

q є1/p.

As є>0 was arbitrary, we see that ∫_A g  d µ=0. As g is positive on A, we conclude that g=0 almost everywhere on A.

A similar argument applied to the set {x∈[0,1] : g(x)≤0} allows us to conclude that g=0 almost everywhere. If K=ℂ, then take real and imaginary parts. □

Proof. If µ=0 then clearly ||µ||=0. If ||µ||=0, then for A∈B(K), let A₁=A, A₂=K∖ A and A₃=A₄=⋯=∅. Then (A_n) is a partition, and so

0 =

∞ ∑ n=1

⎪ ⎪

µ(An)

⎪ ⎪

=

⎪ ⎪

µ(A)

⎪ ⎪

+

⎪ ⎪

µ(K∖ A)

⎪ ⎪

.

Hence µ(A)=0, and so as A was arbitrary, we have that µ=0.

Clearly ||aµ|| = | a |||µ|| for a∈K and µ∈ M_K(K).

For µ,λ∈ M_K(K) and a partition (A_n), we have that

∑ n

⎪ ⎪

(µ+λ)(An)

⎪ ⎪

=

∑ n

⎪ ⎪

µ(An)+λ(An)

⎪ ⎪

≤

∑ n

⎪ ⎪

µ(An)

⎪ ⎪

+

∑ n

⎪ ⎪

λ(An)

⎪ ⎪

≤

⎪⎪ ⎪⎪

µ

⎪⎪ ⎪⎪

+

⎪⎪ ⎪⎪

λ

⎪⎪ ⎪⎪

.

As (A_n) was arbitrary, we see that ||µ+λ|| ≤ ||µ|| + ||λ||. □

Proof. See a book on (point set) topology. □

Proof. If 0≤ f≤χ_U, then

0 =

∫

K

0  d µ ≤

∫

K

f  d µ ≤

∫

K

χU  d µ = µ(U).

Conversely, let F=K∖ U, a closed set. Let E⊆ U be closed. By Urysohn Lemma 26, there exists f:K→[0,1] continuous with f(E)={1} and f(F)={0}. So χ_E ≤ f ≤ χ_U, and hence

µ(E) ≤

∫

K

f  d µ ≤ µ(U).

As µ is regular,

µ(U) = sup

⎧ ⎨ ⎩

µ(E) : E⊆ U closed

⎫ ⎬ ⎭

≤ sup

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

∫

K

f  d µ : 0≤ f≤χU

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

≤ µ(U).

Hence we have equality throughout. □

Proof. Let (A,B) be a Hahn decomposition (Thm. 36) for µ. For f∈ C_ℝ(K) with ||f||_∞≤ 1, we have that

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∫   K  f  d µ  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ≤  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∫   A  f  d µ  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∫   B  f  d µ  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ =  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∫   A  f  d µ+  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ∫   B  f  d µ−  ⎪ ⎪ ⎪ ⎪ ⎪ ⎪                       ≤  ∫   A   ⎪ ⎪ f  ⎪ ⎪   d µ+ +  ∫   B   ⎪ ⎪ f  ⎪ ⎪   d µ− ≤  ⎪⎪ ⎪⎪ f ⎪⎪ ⎪⎪ ∞ ⎛ ⎝ µ(A) − µ(B) ⎞ ⎠ ≤  ⎪⎪ ⎪⎪ f ⎪⎪ ⎪⎪ ∞  ⎪⎪ ⎪⎪ µ ⎪⎪ ⎪⎪ ,

using the fact that µ(B)≤0 and that (A,B) is a partition of K.

Conversely, as µ is regular, for є>0, there exist closed sets E and F with E⊆ A, F⊆ B, and with µ₊(E)> µ₊(A)−є and µ₋(F)>µ₋(B)−є. By Urysohn Lemma 26, there exists f:K→[0,1] continuous with f(E)={1} and f(F)={0}. Let g=2f−1, so g is continuous, g takes values in [−1,1], and g(E)={1}, g(F)={−1}. Then

∫   K  g  d µ =  ∫   E  1  d µ +  ∫   F  −1  d µ +  ∫   K∖ (E⋃ F)  g  d µ                      = µ(E) − µ(F) +  ∫   A∖ E  g  d µ +  ∫   B∖ F  g  d µ

As E⊆ A, we have µ(E) = µ₊(E), and as F⊆ B, we have −µ(F)=µ₋(F). So

∫   K  g  d µ > µ+(A)−є + µ−(B) − є +  ∫   A∖ E  g  d µ +  ∫   B∖ F  g  d µ                      ≥  ⎪ ⎪ µ(A)  ⎪ ⎪  +  ⎪ ⎪ µ(B)  ⎪ ⎪  − 2є −  ⎪ ⎪ µ(A∖ E)  ⎪ ⎪  −  ⎪ ⎪ µ(B∖ F)  ⎪ ⎪                       ≥  ⎪ ⎪ µ(A)  ⎪ ⎪  +  ⎪ ⎪ µ(B)  ⎪ ⎪  − 4є.

As є>0 was arbitrary, we see that ||φ_µ|| ≥ | µ(A) |+| µ(B) |=||µ||. □

Proof. For any function f such that ||f||_∞≤ 1 the function 1−f is non negative thus: F(1)−F(f)=F(1−f)>0, Thus F(1)>F(f), that is F is bounded and its norm is F(1). □

Proof. First, for f∈ C_ℝ(K) with f≥0, we define

λ+(f) = sup ⎧ ⎨ ⎩ λ(g) : g∈ Cℝ(K), 0≤ g≤ f  ⎫ ⎬ ⎭ ≥0,                    λ−(f) = λ+(f) − λ(f) = sup ⎧ ⎨ ⎩ λ(g)−λ(f): g∈ Cℝ(K), 0≤ g≤ f  ⎫ ⎬ ⎭                     = sup ⎧ ⎨ ⎩ λ(h): h∈ Cℝ(K), 0≤ h+f≤ f  ⎫ ⎬ ⎭                     = sup ⎧ ⎨ ⎩ λ(h): h∈ Cℝ(K), −f ≤ h ≤ 0  ⎫ ⎬ ⎭ ≥ 0.

In a sense, this is similar to the Hahn decomposition (Thm. 36).

We can check that

λ+(tf) = tλ+(f),   λ−(tf) = tλ−(f)    (t≥0, f≥0).

For f₁,f₂≥ 0, we have that

λ+(f1+f2) = sup ⎧ ⎨ ⎩ λ(g): 0≤ g ≤ f1+f2  ⎫ ⎬ ⎭                     = sup ⎧ ⎨ ⎩ λ(g1+g2): 0≤ g1+g2 ≤ f1+f2  ⎫ ⎬ ⎭                     ≥ sup ⎧ ⎨ ⎩ λ(g1) + λ(g2): 0≤ g1≤ f1, 0 ≤ g2 ≤ f2  ⎫ ⎬ ⎭                     = λ+(f1) + λ+(f2).

Conversely, if 0≤ g≤ f₁+f₂, then set g₁ = min(g,f₁), so 0≤ g₁ ≤ f₁. Let g₂ = g−g₁ so g₁≤ g implies that 0≤ g₂. For x∈ K, if g₁(x)=g(x) then g₂(x) = 0 ≤ f₂(x); if g₁(x)=f₁(x) then f₁(x)≤ g(x) and so g₂(x) = g(x)−f₁(x) ≤ f₂(x). So 0 ≤ g₂ ≤ f₂, and g = g₁ + g₂. So in the above displayed equation, we really have equality throughout, and so λ₊(f₁+f₂) = λ₊(f₁) + λ₊(f₂). As λ is additive, it is now immediate that λ₋(f₁+f₂) = λ₋(f₁) + λ₋(f₂)

For f∈ C_ℝ(K) we put f₊(x)=max(f(x),0) and f₋(x)=−min(f(x),0). Then f_±≥ 0 and f=f₊−f₋. We define:

λ+(f) = λ+(f+) − λ+(f−),   λ−(f) = λ−(f+) − λ−(f−).

As when we were dealing with integration, we can check that λ₊ and λ₋ become linear functionals; by the previous Lemma they are bounded. □

Proof. Let us show uniqueness. If µ₁,µ₂∈ M_K(K) both induce λ then µ = µ₁−µ₂ induces the zero functional on C_K(K). So for f∈ C_ℝ(K),

0 = ℜ  ∫   K  f  d µ =  ∫   K  f  d µr                      = ℑ ∫   K  f  d µ =  ∫   K  f  d µi.

So µ_r and µ_i both induce the zero functional on C_ℝ(K). By Lemma 28, this means that ||µ_r|| = ||µ_i||=0, showing that µ = µ_r + iµ_i = 0, as required.

Existence is harder, and we shall only sketch it here. Firstly, we shall suppose that K=ℝ and that λ is positive.

Motivated by the above Lemmas 27 and 28, for U⊆ K open, we define

µ*(U) = sup

⎧ ⎨ ⎩

λ(f):  f∈ Cℝ(K),       0≤ f≤χU,  supp(f)⊆ U

⎫ ⎬ ⎭

.

For A⊆ K general, we define

µ*(A) = inf

⎧ ⎨ ⎩

µ*(U):  U⊆ K  is open,       A⊆ U

⎫ ⎬ ⎭

.

We then proceed to show that

• µ^* is an outer measure: this requires a technical topological lemma, where we make use of the support condition in the definition.
• We then check that every open set in µ^*-measurable.
• As B(K) is generated by open sets, and the collection of µ^*-measurable sets is a σ-algebra, it follows that every member of B(K) is µ^*-measurable.
• By using results from Section 12, it follows that if we let µ be the restriction of µ^* to B(K), then µ is a measure on B(K).
• We then check that this measure is regular.
• Finally, we show that µ does induce the functional λ. Arguably, it is this last step which is the hardest (or least natural to prove).

If λ is not positive, then by Lemma 31 represent it as λ=λ₊−λ₋ for positive λ_±. As λ₊ and λ₋ are positive functionals, we can find µ₊ and µ₋ positive measures in M_ℝ(K) such that

λ+(f) =

∫

K

f  d µ+,   λ−(f) =

∫

K

f  d µ−    (f∈ Cℝ(K)).

Then if µ = µ₊ − µ₋, we see that

λ(f) = λ+(f) − λ−(f) =

∫

K

f  d µ    (f∈ Cℝ(K)).

Finally, if K=ℂ, then we use the same “complexification” trick from the proof of the Hahn-Banach Theorem 15. Namely, let λ∈ C_ℂ(K)^*, and define λ_r, λ_i∈ C_ℝ(K)^* by

λr(f) = ℜ  λ(f),   λi(f) = ℑ λ(f)    ( f∈ Cℝ(K) ).

These are both clearly ℝ-linear. Notice also that | λ_r(f) | = | ℜ λ(f) | ≤ | λ(f) | ≤ ||λ|| ||f||_∞, so λ_r is bounded; similarly λ_i.

By the real version of the Riesz Representation Theorem, there exist charges µ_r and µ_i such that

ℜ λ(f) = λr(f) =

∫

K

f  d µr,   ℑλ(f) = λi(f) =

∫

K

f  d µi    (f∈ Cℝ(K) ).

Then let µ=µ_r+iµ_i, so for f∈ C_ℂ(K),

∫   K  f  d µ =  ∫   K  f  d µr + i ∫   K  f  d µi                      =  ∫   K  ℜ (f)  d µr + i ∫   K  ℑ(f)  d µr  + i ∫   K  ℜ (f)  d µi −  ∫   K  ℑ(f)  d µi                      = λr(ℜ (f)) + iλr(ℑ(f)) + iλi(ℜ (f)) − λi(ℑ(f))                      = ℜ  λ(ℜ (f)) + iℜ λ(ℑ(f)) +iℑλ(ℜ (f)) − ℑλ(ℑ(f))                      = λ(ℜ (f) + iℑ(f)) = λ(f),

as required. □