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Measure and Integration, Fall 2008
Lecture 2
August 7, 2008
1
1.1
Measure Spaces
Theorem
Let f : Ω → R and let f −1 ([−∞, α)) ∈ A∀α ∈ R. Then f is measurable.
1.1.1
Proof
∞
∞
[
[
Consider f −1 ((α, ∞]) = f −1 ( [α + n1 , ∞])c =
f −1 ([α + n1 , ∞])c =
i=1
i=1
S∞
1
−1
c
(f
[α
+
,
∞])
which
is
measurable
i=1
n
let α < β ∈ R
Then f −1 ((α, β)) = f −1 ([−∞, β) ∩ (α, ∞]) and thus f is measurable.
1.2
Lemma
Suppose (Ω, A) is a measurable space, X, Y are topological spaces. Let g : X → Y be continuous and let
f : Ω → X be measurable. Then g ◦ f is measurable.
Proof
Let V ⊂ Y be open. As g is continuous, U = g −1 (V ) is open and as f is measurable, E = f −1 (U ) is
measurable. Now (g ◦ f )(E) = V and hence, the preimage of every open set in Y is measurable in Ω
1.3
Lemma
Let (Ω, A) be a measurable space and let u, v : Ω → R be measurable functions. Let Y be a topological
space and let f : R2 → Y be continuous. Then the function Φ : Ω → Y defined by Φ(x) = f (u(x), v(x)) is
measurable.
Proof
Let V be open in Y . f being continuous, f −1 (V ) is open in R2 . As R2 is second-countable, f −1 (V ) can be
∞
[
represented as countable union of open rectangles. Hence, f −1 (V ) =
Ui where Ui = (ai , bi ) × (ci , di ). As
i=1
u, v are measurable, u−1 ((ai , bi )) ∩ v −1 ((ci , di )) is measurable for any i. As f −1 (V ) can be represented as
countable union of Ui , its preimage in X is measurable. Hence, Φ(x) = f (u(x), v(x)) is measurable.
1.4
Theorem
Let (Ω, A) be a measurable space. Let L be the space of all measurable functions f : Ω → R. Then L is a
vector space over R, and is closed under multiplication.
Proof
It follows from Lemma 1.3 that (f + g)(x), (f · g)(x), af (x) are all measurable whenever f, g are measurable.
The other properties follow immediately.
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1.5
Corollary
If u, v are real measurable functions, the function f = u + iv is complex measurable.
1.6
Corollary
If f = u + iv is a complex measurable function (where u, v are real valued), then u, v, |f | are all measurable.
1.7
Corollary
Let E ∈ A be(measurable. The characteristic function of χE : Ω → R is defined as
1 If x ∈ E
χE (x) =
if x ∈ E; χE (x) = 0 otherwise.
0 If x ∈ E c
Then, the characteristic function is measurable
1.8
Corollary
Let f be a complex measurable function. Then there exists a complex measurable function α such that
|α(x)| = 1∀x and f = α|f |
Proof
Consider E = {x : f (x) = 0}. We claim that E is measurable, for C \ {0} being open, K = f −1 (C \ {0}) is
measurable and hence, E = K c is measurable.
E (z)
Hence, the characteristic function χE is measurable. Now define the function α as α(z) = |ff (z)+χ
(z)|+χE (z)
2
Borel algebra
2.1
Definition
Let (X, T ) be a topological space. Then σ(T ) is called the Borel σ-field B. Members of the Borel σ-field
are called Borel sets
(A countable union of closed sets is called Fσ whereas a countable intersection of open sets is called Gδ .
Then all Fσ and Gδ are members of B)
(X, B) is called a Borel-measureable space.
2.2
Definition
Let (X, B) be a Borel measureable space, let Y be a topological space and let f : X → Y . f is said to be a
Borel measurable function iff f −1 (V ) is Borel measurable for every open set V .
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Exercises
Let (Ω, A) be a measurable space, let Y be a topological space and let f : X → Y be any map. Then,
1. M = {E ⊂ Y : f −1 (E) ∈ A} is a σ-field in Y
2. If E is a Borel measurable subset of Y and if f is measurable then f −1 (E) ∈ A
3. If f is measurable and Z s a topological space, g : Y → Z is Borel measurable, then g ◦ f is measurable.
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3.1
Solutions
1. Let E ∈ M. Then f −1 (E) ∈ A, and hence(f −1 (E))c ∈ A implying that f −1 (E c ) ∈ A
Similarly, let E1 , E2 ∈ M hence f −1 (E1 ) ∩ f −1 (E2 ), f −1 (E1 ) ∪ f −1 (E2 ) ∈ A that is, E1 ∩ E2 , E1 ∪ E2 ∈
M
∞
∞
\
\
Now, let E1 , E2 , . . . ∈ M. Then
f −1 (Ei ) ∈ A and hence,
Ei ∈ M. Similarly, we can show that
i=1
i=1
M is closed under countable union.
2. We already know that if C is a collection of subsets of Y , every element of σ(C) is either a member of
C or is a countable union, intersection or complement of members of C. Hence, by the previous result,
f −1 (E) is measurable for every Borel set E.
3. Let V ⊂ Z be open. Then, g −1 (V ) is Borel measurable and by previous result, f −1 (g −1 (V )) is
measurable. Thus, g ◦ f is a measurable function.
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