19.5. PARTITIONS OF UNITY 511

Proof: Let ReA ≡ {Re f : f ∈A }, ImA ≡{Im f : f ∈A }.Claim 1: ReA = ImAProof of claim: A typical element of ReA is Re f where f ∈ A , then from Lemma

19.4.4, Re( f ) = Im(i f̄)∈ ImA . Thus ReA ⊆ ImA . By assumption, i f̄ ∈A . The other

direction works the same. Just use the other formula in Lemma 19.4.4.Claim 2: Both ReA and ImA are real algebras.Proof of claim: It is obvious these are both real vector spaces. Since these are equal, it

suffices to consider ReA . It remains to show that ReA is closed with respect to products.

f + f̄2

g+ ḡ2

=14[

f g+ f ḡ+ f̄ g+ f g]=

14[2Re( f g)+2Re

(f̄ g)]

Now by assumption, f g ∈A and so Re( f g) ∈ ReA . Also Re(

f̄ g)∈ ReA because both

f̄ ,g are in A and it is an algebra. Thus, the above is in ReA because, as noted, this is areal vector space.

Claim 3: A = ReA + i ImAProof of claim: If f ∈ A , then f = f+ f̄

2 + i f− f̄2i ∈ ReA + i ImA so A ⊆ ReA +

i ImA . Now a generic element of ReA + i ImA is Re( f )+ i Im(g) for f ,g ∈A .

Re( f )+ i Im(g)≡ f + f̄2

+ i(

g− ḡ2i

)=

f +g2

+f̄ − ḡ

2∈A

because A is closed with respect to conjugates. Thus ReA + i ImA ⊆A .Both ReA and ImA must separate the points. Here is why: If x1 ̸= x2, then there exists

f ∈A such that f (x1) ̸= f (x2) . If Im f (x1) ̸= Im f (x2) , this shows there is a function inImA , Im f which separates these two points. If Im f fails to separate the two points, thenRe f must separate the points and so, by Lemma 19.4.4,

Re f (x1) = Im(i f̄ (x1)

)̸= Re f (x2) = Im

(i f̄ (x2)

)Thus ImA separages the points. Similarly ReA separates the points using a similar argu-ment or because it is equal to ImA .

Neither ReA nor ImA annihilate any point. This is easy to see because if x is apoint, there exists f ∈ A such that f (x) ̸= 0. Thus either Re f (x) ̸= 0 or Im f (x) ̸= 0. IfIm f (x) ̸= 0, this shows this point is not annihilated by ImA . Since they are equal, ReAdoes not annihilate this point either.

It follows from Theorem 19.4.3 that ReA and ImA are dense in the real valued func-tions of C0 (X). Let f ∈C0 (X) . Then there exists {hn} ⊆ReA and {gn} ⊆ ImA such thathn→ Re f uniformly and gn→ Im f uniformly. Therefore, hn + ign ∈A and it convergesto f uniformly. ■

19.5 Partitions of UnityAs before, the idea of a partition of unity if of fundamental significance. It will be used toconstruct measures.

Definition 19.5.1 Define spt( f ) (support of f ) to be the closure of the set {x :f (x) ̸= 0}. If V is an open set, Cc(V ) will be the set of continuous functions f , definedon Ω having spt( f )⊆V . Thus in Theorem 19.1.23, f ∈Cc(V ).

19.5. PARTITIONS OF UNITY 511Proof: Let Re” = {Ref: f€ #}, Im# ={Imf: fe}.Claim 1: Rev =Im#Proof of claim: A typical element of Re. is Re f where f € &, then from Lemma19.4.4, Re(f) =Im(if) € Im. Thus Rew C Im”. By assumption, if € &/. The otherdirection works the same. Just use the other formula in Lemma 19.4.4.Claim 2: Both Re. and Im /& are real algebras.Proof of claim: It is obvious these are both real vector spaces. Since these are equal, itsuffices to consider Re .&/. It remains to show that Re. is closed with respect to products.prFete_a) 7 lfg+fe+Fe-+ Fal = 5 [2Re(fe) +2Re (Fa)]Now by assumption, fg € / and so Re(fg) € Re”. Also Re (fg) € Re. because bothf,g are in & and it is an algebra. Thus, the above is in Re & because, as noted, this is areal vector space.Claim 3: o =Rew# +ilma# - 7Proof of claim: If f € ./, then f = ir +ifff €ReW+ilme so PH CReW +iIm.&. Now a generic element of Re +iIm&# is Re(f) +ilm(g) for f,g € &.Re(f) +itm(g) = 25 4() = fs begbecause . is closed with respect to conjugates. Thus Re’ +iImaw# C &,Both Re. Y and Im#/ must separate the points. Here is why: If x, 4 x2, then there existsf € & such that f (x1) 4 f (x2). If Imf (x1) 4 Im f (x2), this shows there is a function inIm.#/, Im f which separates these two points. If Im f fails to separate the two points, thenRe f must separate the points and so, by Lemma 19.4.4,Re f (x1) =Im (if (x1)) Re f (x2) = Im (if (x2))Thus Im .& separages the points. Similarly Re. separates the points using a similar argu-ment or because it is equal to Im.¢%.Neither Re. nor Im.# annihilate any point. This is easy to see because if x is apoint, there exists f € o/ such that f (x) 4 0. Thus either Re f (x) 4 0 or Im f (x) 4 0. IfIm f (x) 4 0, this shows this point is not annihilated by Im.%. Since they are equal, Re /does not annihilate this point either.It follows from Theorem 19.4.3 that Re.e” and Im.& are dense in the real valued func-tions of Co (X). Let f € Co (X). Then there exists {h,} C Re. & and {g,} C Im” such thathy + Ref uniformly and g, — Im f uniformly. Therefore, 4, +ig, € & and it convergesto f uniformly.19.5 Partitions of UnityAs before, the idea of a partition of unity if of fundamental significance. It will be used toconstruct measures.Definition 19.5.1 Define spt(f) (support of f) to be the closure of the set {x :f(x) 40}. If V is an open set, C.(V) will be the set of continuous functions f, definedon Q having spt(f) C V. Thus in Theorem 19.1.23, f € C-(V).