Download Math 641 Lecture #2 ¶1.9,1.11,1.12

Math 641 Lecture #2
¶1.9,1.11,1.12
Consequences (1.9) (of Theorems 1.7 and 1.8).
a) The function Φ : R2 → C defined by Φ(u, v) = u + iv is continuous, where we take the
usual topology for C.
So if u, v : X → R are measurable, then f (x) = Φ(u(x), v(x)) = u(x) + iv(x) is a
complex measurable function on X.
b) The functions
Re : C → R defined by u + iv 7→ u,
Im : C → R defined by u + iv 7→ √
v, and
| · | : C → R defined by u + iv 7→ u2 + v 2
are continuous.
For a complex measurable f : X → C defined by f = u + iv, the compositions
Re(f ) = u,
Im(f )√= v, and
|f | = u2 + v 2
are measurable.
c) The functions
Φ : R2 → R defined by Φ(s, t) = s + t and
Ψ : R2 → R defined by Ψ(s, t) = st
are continuous.
For complex measurable functions f, g on X, the compositions
Φ(f (x), g(x)) = f (x) + g(x) = (f + g)(x) and
Ψ(f (x), g(x)) = f (x)g(x) = (f g)(x)
are complex measurable.
d) The characteristic function of a measurable set E in X is defined by
(
1 if x ∈ E
χE (x) =
0 if x ∈
/E
It is a complex measurable function because for any open V in C,

∅
if 1 ∈
/ V and 0 ∈
/ V,



E
if 1 ∈ V but 0 ∈
/ V,
χ−1
E (V ) = {x ∈ X : χE (x) ∈ V } =

X
if 0, 1 ∈ V,


 C
E
if 0 ∈ V but 1 ∈
/ V,
is measurable in X.
e) If f is a complex measurable function on X, then there is a complex measurable
function α on X such that |α| = 1 and f = α|f |.
Proof: Let Y = C \ {0}, and set ϕ(z) = z/|z| , z = u + iv.
Let E = {x ∈ X : f (x) = 0}, and set α(x) = ϕ f (x) + χE (x) , x ∈ X.
If x ∈ E, then
α(x) =
0+1
= 1;
|0 + 1|
If x ∈
/ E, then
α(x) =
f (x) + 0
f (x)
=
.
|f (x) + 0|
|f (x)|
Thus |α(x)| = 1 for all x ∈ X, and
(
0 = α(x)|f (x)| if x ∈ E
f (x) =
α(x)|f (x)|
if x ∈
/ E.
The function ϕ : Y → C is continuous, and the set
E = f −1 ({0})
!
∞
\
1
= f −1
{z ∈ C : |z| < }
n
n=1
∞
\
1
−1
=
f
{z ∈ C : |z| < }
n
n=1
[open disks centered at origin with radius 1/n]
[set theoretic property of preimages and intersections]
is measurable.
Thus, since the sum of two measurable functions is measurable, we have α(x) is measurable.
Definition (1.11). Let τ be a topology on X. The smallest σ-algebra in X that contains
τ is the Borel σ-algebra in X, and is denoted by B or BX . The members of B are the
Borel sets of X.
Consequences.
a) Since each open set in X is a Borel set, the countable intersection of open sets is a
Borel set and is called a Gδ .
b) Each closed set in X is a Borel set because the complement of a closed set is open,
and so the countable union of closed sets is a Borel set and is called an Fσ .
Definition. Let X, Y be topological spaces. A function f : X → Y is a Borel function
(or Borel measurable) if
f −1 (V ) ∈ BX for all open V in Y.
Observation. A continuous f : X → Y is Borel measurable because for every open V
in Y , f −1 (V ) is open in X, hence f −1 (V ) ∈ BX .
Lemma (1.12a). If M is a σ-algebra in X, and f : X → Y for a topological space Y ,
then
Ω = {E ⊂ Y : f −1 (E) ∈ M}
is a σ-algebra in Y .
Proof: The set Y belongs to Ω because f −1 (Y ) = X ∈ M.
For A ∈ Ω, f −1 (A) ∈ M. So AC ∈ Ω because f −1 (AC ) = [f −1 (A)]C ∈ M.
For An ∈ Ω, n = 1, 2, 3, . . ., we have f −1 (An ) ∈ M, n = 1, 2, 3, . . ..
−1
∞
−1
(An ) ∈ M.
(∪∞
So ∪∞
n=1 An ) = ∪n=1 f
n=1 An ∈ Ω because f
Thus, Ω is a σ-algebra in Y .
Theorem (1.12b,c). Let M be a σ-algebra in X, let Y be a topological space, and let
f : X → Y.
i) If f is measurable and E is a Borel set in Y , then f −1 (E) ∈ M.
ii) If Y = [−∞ , ∞] and f −1 (α , ∞] ∈ M for all α ∈ R, then f is measurable.
Proof: i) Let Ω = {E ⊂ Y : f −1 (E) ∈ M}
Measurability of f implies that Ω contains all the open sets in Y . (f −1 (V ) ∈ M for all
open V in Y )
By the Lemma, Ω is a σ-algebra; hence BY ⊂ Ω. (BY is the smallest σ-algebra that
contains the open sets in Y .)
In other words, f −1 (E) ∈ M for every E ∈ BY .
ii) Let Ω = {E ⊂ [−∞, ∞] : f −1 (E) ∈ M}.
By the Lemma, Ω is a σ-algebra in [−∞, ∞].
Choose α ∈ R, and a sequence {αn } ∈ R such that αn < α and αn → α.
Since f −1 (αn , ∞] ∈ M, it follows that
[−∞, α) =
∞
[
[−∞, αn ] =
n=1
∞
[
(αn , ∞]C ∈ Ω
n=1
because Ω is a σ-algebra and because (αn , ∞] ∈ Ω.
Hence, (α, β) = [−∞, β) ∩ (α, ∞] ∈ Ω.
Every open set in [−∞, ∞] is a countable union of intervals of the types (α, ∞], [−∞, β),
and (α, β).
Thus, B[−∞,∞] ⊂ Ω, implying that f is measurable, i.e, f −1 (V ) ∈ M for all open V in
[−∞, ∞].
Corollary (1.12(d)). If f : X → Y is measurable and g : Y → Z is Borel measurable,
then h = g ◦ f : X → Z is measurable.
Proof: Let V be open in Z. Then g −1 (V ) is a Borel set in Y .
f −1 g −1 (V ) , the Theorem (part i) shows that h−1 (V ) ∈ M.
Since h−1 (V ) =