Rudin's Real and Complex Analysis - up to Integration review:
1. A positive measure is a function defined on a σ-algebra R whose ramge is [0,∞] and which is countably additive. This means that if Ai is disjoint countable collection of members of R, then
μ(∪∞i=1Ai)=∞∑i=1μ(Ai)
Here the assumption is that μ(Ai)m<∞ for at least one A∈R.
2. A measure space is a measurable space which has a postive measure defined on the σ-algebra of its measurable sets.
3. A complex measure is a complex-valued countable additive function defined on a σ-algebra.
Theorem 1.19:
Let μ be the positive measure on the σ-algebra on R. Then
a. μ(ϕ)=0.
b. μ(A1∪A2∪⋯∪An)=μ(A1)+μ(A2)+⋯+μ(An) if A1,A2,⋯,An are pairwise disjoint members of R.
c. A⊂B implies μ(A)≤μ(B) if A,B∈R.
d. μ(An)→μ(A) as n→∞ if A=∪∞n=1An,An∈R and A1⊂A2⊂⋯
e. A⊂B implies μ(A)≤μ(B) if A,B∈R.
d. μ(An)→μ(A) as n→∞ if A=∩∞n=1An,An∈R and A1⊃A2⊃A3⋯
All these properties with exception of c hold for complex measure. (b) is called finite additivity and (c) is called monotonicity.
Proofs are simple.
For (a)
Starts off by letting A∈R so that μ(A)<∞ and take A1=A and A2=A3=⋯=ϕ. Use the observation that μ(cup∞i=1Ai)=σ∞i=1μ(Ai).
μ(A1∪ϕ)=μ(A1)+μ(ϕ)
μ(A1)=μ(A1)+μ(ϕ)
Hence μ(ϕ)=0.
For (b), taking An+1=An+2=⋯=ϕ will lead to
μ(∪∞i=1)=μ(∪ni=1=μ(A1)+μ(A2)+⋯+μ(An)
For (c)
Since B=A∪(B−A) and A∩(B−A)=ϕ. Later shows that A and (B−A) are disjoint.
For (d), put B1=A1 and put Bn=An−An−1 for n=2,3,4,⋯. Then Bn∈R follows from definition of measurability. Clearly Bi∪Bj=ϕ if i≠j and this has the required disjointedness, An=B1∪B2∪⋯∪Bn and A=∪i=1∞Bi. Hence,
μ(An)=n∑i=1μ(Bi)
(d) follows from definition of sum of infinite series.
Similar proof for (e).
Finally,an example pops up. Notice till this time, book proceeds with no examples. Example illustrates counting measure and unit mass concentrated at a point.
Then there is a nice excursion on the terminology.
1. A positive measure is a function defined on a σ-algebra R whose ramge is [0,∞] and which is countably additive. This means that if Ai is disjoint countable collection of members of R, then
μ(∪∞i=1Ai)=∞∑i=1μ(Ai)
Here the assumption is that μ(Ai)m<∞ for at least one A∈R.
2. A measure space is a measurable space which has a postive measure defined on the σ-algebra of its measurable sets.
3. A complex measure is a complex-valued countable additive function defined on a σ-algebra.
Theorem 1.19:
Let μ be the positive measure on the σ-algebra on R. Then
a. μ(ϕ)=0.
b. μ(A1∪A2∪⋯∪An)=μ(A1)+μ(A2)+⋯+μ(An) if A1,A2,⋯,An are pairwise disjoint members of R.
c. A⊂B implies μ(A)≤μ(B) if A,B∈R.
d. μ(An)→μ(A) as n→∞ if A=∪∞n=1An,An∈R and A1⊂A2⊂⋯
.
e. A⊂B implies μ(A)≤μ(B) if A,B∈R.
d. μ(An)→μ(A) as n→∞ if A=∩∞n=1An,An∈R and A1⊃A2⊃A3⋯
and μ(Ai) is finite..
All these properties with exception of c hold for complex measure. (b) is called finite additivity and (c) is called monotonicity.
Proofs are simple.
For (a)
Starts off by letting A∈R so that μ(A)<∞ and take A1=A and A2=A3=⋯=ϕ. Use the observation that μ(cup∞i=1Ai)=σ∞i=1μ(Ai).
μ(A1∪ϕ)=μ(A1)+μ(ϕ)
μ(A1)=μ(A1)+μ(ϕ)
.
Hence μ(ϕ)=0.
For (b), taking An+1=An+2=⋯=ϕ will lead to
μ(∪∞i=1)=μ(∪ni=1=μ(A1)+μ(A2)+⋯+μ(An)
.
For (c)
Since B=A∪(B−A) and A∩(B−A)=ϕ. Later shows that A and (B−A) are disjoint.
For (d), put B1=A1 and put Bn=An−An−1 for n=2,3,4,⋯. Then Bn∈R follows from definition of measurability. Clearly Bi∪Bj=ϕ if i≠j and this has the required disjointedness, An=B1∪B2∪⋯∪Bn and A=∪i=1∞Bi. Hence,
μ(An)=n∑i=1μ(Bi)
and μ(A)=∑∞i=1μ(Bi).
(d) follows from definition of sum of infinite series.
Similar proof for (e).
Finally,an example pops up. Notice till this time, book proceeds with no examples. Example illustrates counting measure and unit mass concentrated at a point.
Then there is a nice excursion on the terminology.
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