Rudin's Real and Complex Analysis - up to Integration review:
1. A positive measure is a function defined on a $\sigma$-algebra $R$ whose ramge is $[0,\infty]$ and which is countably additive. This means that if $A_i$ is disjoint countable collection of members of $R$, then
$$\mu(\cup_{i=1}^\infty A_i ) = \sum_{i=1}^{\infty} \mu(A_i)$$
Here the assumption is that $\mu(A_i)m < \infty$ for at least one $A \in R$.
2. A measure space is a measurable space which has a postive measure defined on the $\sigma$-algebra of its measurable sets.
3. A complex measure is a complex-valued countable additive function defined on a $\sigma$-algebra.
Theorem 1.19:
Let $\mu$ be the positive measure on the $\sigma$-algebra on $R$. Then
a. $\mu(\phi) = 0 $.
b. $\mu(A_1 \cup A_2 \cup \cdots \cup A_n) = \mu(A_1)+\mu(A_2)+\cdots+\mu(A_n)$ if $A_1,A_2,\cdots,A_n$ are pairwise disjoint members of $R$.
c. $A \subset B$ implies $\mu(A) \leq \mu(B)$ if $A,B \in R$.
d. $\mu(A_n) \rightarrow \mu(A)$ as $n \rightarrow \infty$ if $A=\cup_{n=1}^\infty A_n, A_n \in R$ and $$A_1 \subset A_2 \subset \cdots$$.
e. $A \subset B$ implies $\mu(A) \leq \mu(B)$ if $A,B \in R$.
d. $\mu(A_n) \rightarrow \mu(A)$ as $n \rightarrow \infty$ if $A=\cap_{n=1}^\infty A_n, A_n \in R$ and $$A_1 \supset A_2 \supset A_3 \cdots$$ and $\mu(A_i)$ 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 $A\in R$ so that $\mu(A) < \infty$ and take $A_1=A$ and $A_2=A_3=\cdots=\phi$. Use the observation that $\mu(cup_{i=1}^\infty A_i) = \sigma_{i=1}^\infty \mu(A_i)$.
$$\mu(A_1 \cup \phi) = \mu(A_1)+\mu(\phi)$$
$$\mu(A_1) = \mu(A_1)+\mu(\phi)$$.
Hence $\mu(\phi) = 0$.
For (b), taking $A_{n+1}=A_{n+2}=\cdots=\phi$ will lead to
$$\mu(\cup_{i=1}^\infty)=\mu(\cup_{i=1}^n=\mu(A_1)+\mu(A_2)+\cdots+\mu(A_n)$$.
For (c)
Since $B=A \cup (B-A)$ and $A \cap (B-A)=\phi$. Later shows that $A$ and $(B-A)$ are disjoint.
For (d), put $B_1=A_1$ and put $B_n = A_n - A_{n-1}$ for $n=2,3,4,\cdots$. Then $B_n \in R$ follows from definition of measurability. Clearly $B_i \cup B_j =\phi$ if $i \neq j$ and this has the required disjointedness, $A_n=B_1 \cup B_2 \cup \cdots \cup B_n$ and $A = \cup{i=1}^\infty B_i$. Hence,
$$\mu(A_n)=\sum_{i=1}^n \mu(B_i)$$ and $\mu(A)=\sum_{i=1}^{\infty}\mu(B_i)$.
(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 $\sigma$-algebra $R$ whose ramge is $[0,\infty]$ and which is countably additive. This means that if $A_i$ is disjoint countable collection of members of $R$, then
$$\mu(\cup_{i=1}^\infty A_i ) = \sum_{i=1}^{\infty} \mu(A_i)$$
Here the assumption is that $\mu(A_i)m < \infty$ for at least one $A \in R$.
2. A measure space is a measurable space which has a postive measure defined on the $\sigma$-algebra of its measurable sets.
3. A complex measure is a complex-valued countable additive function defined on a $\sigma$-algebra.
Theorem 1.19:
Let $\mu$ be the positive measure on the $\sigma$-algebra on $R$. Then
a. $\mu(\phi) = 0 $.
b. $\mu(A_1 \cup A_2 \cup \cdots \cup A_n) = \mu(A_1)+\mu(A_2)+\cdots+\mu(A_n)$ if $A_1,A_2,\cdots,A_n$ are pairwise disjoint members of $R$.
c. $A \subset B$ implies $\mu(A) \leq \mu(B)$ if $A,B \in R$.
d. $\mu(A_n) \rightarrow \mu(A)$ as $n \rightarrow \infty$ if $A=\cup_{n=1}^\infty A_n, A_n \in R$ and $$A_1 \subset A_2 \subset \cdots$$.
e. $A \subset B$ implies $\mu(A) \leq \mu(B)$ if $A,B \in R$.
d. $\mu(A_n) \rightarrow \mu(A)$ as $n \rightarrow \infty$ if $A=\cap_{n=1}^\infty A_n, A_n \in R$ and $$A_1 \supset A_2 \supset A_3 \cdots$$ and $\mu(A_i)$ 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 $A\in R$ so that $\mu(A) < \infty$ and take $A_1=A$ and $A_2=A_3=\cdots=\phi$. Use the observation that $\mu(cup_{i=1}^\infty A_i) = \sigma_{i=1}^\infty \mu(A_i)$.
$$\mu(A_1 \cup \phi) = \mu(A_1)+\mu(\phi)$$
$$\mu(A_1) = \mu(A_1)+\mu(\phi)$$.
Hence $\mu(\phi) = 0$.
For (b), taking $A_{n+1}=A_{n+2}=\cdots=\phi$ will lead to
$$\mu(\cup_{i=1}^\infty)=\mu(\cup_{i=1}^n=\mu(A_1)+\mu(A_2)+\cdots+\mu(A_n)$$.
For (c)
Since $B=A \cup (B-A)$ and $A \cap (B-A)=\phi$. Later shows that $A$ and $(B-A)$ are disjoint.
For (d), put $B_1=A_1$ and put $B_n = A_n - A_{n-1}$ for $n=2,3,4,\cdots$. Then $B_n \in R$ follows from definition of measurability. Clearly $B_i \cup B_j =\phi$ if $i \neq j$ and this has the required disjointedness, $A_n=B_1 \cup B_2 \cup \cdots \cup B_n$ and $A = \cup{i=1}^\infty B_i$. Hence,
$$\mu(A_n)=\sum_{i=1}^n \mu(B_i)$$ and $\mu(A)=\sum_{i=1}^{\infty}\mu(B_i)$.
(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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