R&C Analysis up to prop 1.25

Rudin's Real and Complex Analysis - Review up to prop 1.25

Main thrust of these sections is to establish Integration of positive functions. All the required machinary for Integration was  established in previous sections.

I felt that section 1.13 and Theorem 1.14 actually should have preceded this section. Only one collollary - The limit of every pointwise convergent sequence of complex measurable functions is measurable is used before this section.

Section 1.13 defines upper limit of sequence $(a_n)$ - called $\beta$ as $$\beta=lim sup_{n \rightarrow \infty} a_n$$. Lower limit gets defined as negative of upper limit. From real analysis, if $(a_n)$ converges, then both upper and lower limits will equal to limit of sequence.

Based on these definitions, sup and inf of sequence of functions are defined.
$(sup_{n}f_n)(x) = sup_n(f_n(x))$
$(lim sup_{n \rightarrow \infty} f_n)(x) = lim sup_{n \rightarrow \infty}(f_n(x))$

And if
$f(x)= lim_{n \rightarrow \infty} f_n(x)$
the limit being assumed to exist at every point $x \in X$, then $f$ is called pointwise limit of sequence.

Then Theorem 1.14 asserts for $f_n:X \rightarrow [-\infty,\infty]$ is measurable function for $n=1,2,3,\cdots$ and
$g = sup_{n \geq 1}f_n, h = lim sup_{n \rightarrow \infty} f_n$
then $g$ and $h$ are measurable.


Proof is simple. Both the corollaries play a critical role in Integration theory.
(a) The limit of every pointwise convergent sequence of complex measurable functions is measurable.
(b) If $f,g$ are measurable with range in $[-\infty,\infty]$, then so are $max{f,g}$ and $min{f,g}$ and in particular, this is true of the functions
$f^{+} = max\{f,0\}$
$f^{-} = min\{f,0\}$
Proof is simple.

Integration of Positive Functions:
Definition starts off the proceedings for a given $R$ - $\sigma$ algebra and $X$ a measurable space. First order of business to  define Integral of simple functions as this leads to Integral of the given function to which simple functions converge pointwise. Definition pops out very nicely from definition of simple function.
If $s:X \rightarrow [0,\infty]$ is measurable simple function of the form
$s = \sum_{i=1}^\infty \alpha_i \chi_{A_i}$ where $\alpha_1,\cdots,\alpha_n$ are distinct values of $s$ and if $E \in R$, simple function integral is defined as
$\int_E s d\mu = \sum_{i=1}^\infty \alpha_i \mu(A_i \cap E)$.

Lebesgue Integral:
Since multiple simple functions may approximate a given function, its integral is defined as
$\int_E f d\mu = sup \int_E s d\mu$.
The sup taken over all measurable simple functions $s$ such that $0 \leq s \leq f$.

Next all the nice properties of Lebesgue Integral are listed as propostions. Functions and sets are assumed to be measurable.

(a) If $0 \leq f \leq g $ then $\int_E f d\mu \leq \int_E g d\mu$
(b) If $A \subset B$ and $f \geq 0$ then $\int_A f d\mu \leq \int_b f d\mu$.
(c) If $ f \geq 0 $ and $c$ is constant, $0 \leq c \lt \infty$, then
$\int_E cf d\mu = c \int_E f d\mu$
(d) If $f(x)=0$ for all $ x \in E$, then $\int_E f d\mu = 0 $ even if $\mu(E) = \infty$.
(e) If $\mu(E)=0$, then $\int_E f d\mu = 0 $ even if $f(x) = \infty$ for every $x \in X$.
(f) If $f \geq 0$, then $\int_E f d\mu = \int_X \chi_E f d\mu$.
Last result is nice as it allows to restrict our definition of integration to integrals over all of $X$, without loosing
generality.

R&C Analysis up to Integration.

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.

R & C Analysis - to 1.17

Rudin's Real and Complex Analysis - up to Section 1.17  review:
 From definition 1.13 to Theorem 1.14 and Corollaries should have been just before Fatau's lemma. Both the definition and Theorem are not used till then. So moving on...

Simple Functions:
Simple function is a complex valued function $s$ on a measurable space $X$ whose range consists of finitely many points.

No examples are given. Perhaps, Rudin assumes that the reader is familiar with such functions. And no motivation is given for such an important concept that leads to abstract integration.

Floor function is an example of simple function. Say we have a range of real numbers from $1$ to $10$. Then the floor function yields integers $1,2,\cdots,10$.

For us, better think of simple functions as step functions. Let $A_i=\{x_i:s(x)=\alpha_i\}$ is a partition of space $X$. Clearly $s$ takes a constant value in each $A_i$, $s\|_{A_i}=\alpha_i$. Using characterstic functions, one can write
$$s = \sum_{i=1}^n \alpha_i \chi_{A_i}$$

Measure based integrals are defined using Simple functions. They play similar role as tiny rectangles in the definition of Riemann integrals.

What is very fascinating is that a sequence of non-decreasing simple functions that converge to positive valued measurable function. That is the Theorem 1.17.

One can test this theorem using some measurable functions. If function $f$ is constant, the theorem is true. If the function describes a line through origin (y=x) for positive x axis, then one can define a simple function at each rational point whose pointwise convergence describes the line through origin.

Theorem 1.17: Let $f:X \rightarrow [0,\infty]$ be measurable. There exists simple functions $s_n$ such that

1. $0 \leq s_1 < \leq s_2<\cdots\leq f_n$.
2. $s_n(x) \rightarrow f(x)$ as $n \rightarrow \infty$ for every $x \in X$.

The proof as given in the text requires a bit of unraveling. Very knotty, indeed.

R&C Analysis to 1.13

Rudin's Real and Complex Analysis - up to Section 1.13  review:

Theorem 1.10 & Borel sets:
Theorem 1.10 establishes a smallest $\sigma$-algebra for any space $X$ formed by any collection of subsets of $X$. The proof is direct and simple. Motivation of smallest sigma algebra becomes clear in next section.

Borel sets are defined in a slightly convoluted way. Simpler definition is in WIKI for Borel sets. $F_\sigma$ and $G_\delta$ sets are defined in terms of countable union of closed sets and countable intersection of open sets. Gives a nice history of this nomenclature.

Every continuous mapping of $X$ is Borel measurable - aka Borel functions.

All these dovetail nicely into Theorem $1.12$

Set up is a space $X$ and a topological space $Y$ and a map $f$ between $X$ into $Y$. Then any collection of sets in $Y$ whose inverse maps belongs to sigma algebra, is a sigma algebra in $Y$. This gives a great way to establish sigma algebras in $Y$.

If $f$ is measurable, any Borel set $E \in Y$ with $f^{-1}(E)$, then $f^{-1}(E) \in \sigma \in X$.

Most applicable proposition (proposition c in the book) is if $Y=[-\infty,\infty]$ and $f^{-1}[(\alpha,\infty)]$ belongs to sigma algebra in $X$, then $f$ is measurable. 

Proofs follow from the definition of measurable spaces.

R&C analysis - 1.9

Rudin's Real and Complex Analysis - Section 1.9 review:

Section $1.9$ deals with consequences of Theorem $1.9$ and $1.8$.
Taking $X$ as a measurable space, a function $f=u+iv$ is measurable when $u,v$ are measurable. This follows by setting
$\Phi(z)=\Phi(u,v)=z=u+iv$ where $\Phi$ is a continuous mapping of a plane into a topological space - $f$ is measurable.

Given $f=u+iv$ is measurable, showing $u$,$v$ are measurable is easy. Assume $Re(z)=u$ be projection of real part. Since $Re(z)$ is continuous, $Re\circ f$ is measurable via Theorem $1.7$. Similarly for $v$ and $|f|$.

Simple proof shows that $f+g$ and $fg$ are measurable for measurable $f,g$.

On a measurable set, definition of characterstic function is given. Then a theorem (point e) on complex measurable function is proved that has no applications for almost several sections!

The construction in this proof is instructive. Given that $f$ is a complex measurable function, need to show there exists a complex measurable function $\alpha$ such that $|\alpha|=1$ and $f=\alpha|f|$. Approach is to use Theorem $1.7$ that requires constructing a continuous function $\phi$ and compose $\phi$ with measurable functions.

Start off by setting $\phi(z)=z/|z|$ for each $z \in Y$ where $Y$ is complex plane with origin removed.

To use characterstic function - a set $E=\{x|f(x)=0\}$ is defined, then

$\alpha(x) = \phi(f(x)+\chi_E)$


Clearly, if $x \in E$, $\chi_E=1$ and $f(x)=0$. Then $\alpha(x)=1$.
If $x \notin E$, then $\chi_E=0$ and $\alpha(x)=\phi(f(x))=f(x)/|f(x)|$. Now Theorem $1.7$ can be applied to prove existence of $\alpha$.

R&C Analysis - up to 1.9

Rudin's Real and Complex Analysis -upto Section 1.9 review:
At this point (if you are following this from previous post, Rudin introduces a Theorem $1.7$ that is fundamental and very useful.

Stated informally,
1. Continuous function of continuous function is continuous.
2. Continuous function of measurable function is measurable.

(2) requires $Y,Z$ are topological spaces for continuous functions  and $X$ is measurable. Of course (1)  requires all spaces be topological.

The proof relies on definitions of continuity and measurability. Measurability requires $f^{-1}(V)$ of any open set $V$ be measurable.

Theorem 1.8 is very puzzling at first. There is no explanation why we need a plane and what it is doing here. Fortunately, reading next section ($1.9$) makes it clear. Also it uses a fact from real analysis - every open set in plane is a countable union of rectangles with sides parallel to axis.

The proof is somewhat complicated - but not too bad. Given $u(x),v(x)$ are measurable, it starts off by setting $f=(u(x),v(x))$ as mapping of $X$ (measurable space) into a plane and setting $h=\Phi\circ f$ and thus requiring measurability of $f$ to complete the proof.  Since $\Phi$ is  defined to be continuous function, the measurability of $f$ completes the proof. To show $f$ is measurable all that's needed to show is $f^{-1}(V)$ for any open set $V$ should be measurable.

R & C Analysis-upto1.4

Rudin's Real and Complex Analysis - upto Section 1.4 review:

The chapter starts off fine with some motivation on why we need better integral than Riemman integral. Very nice observation in this section is - "The passage from Riemann's theory of integration to that Lesbegue is a process of completion...It is of same fundamental importance in analysis as is the construction of real number system from rationals".

Next section is a simple review of set theory. Then the concept of measurability is introduced while pointing out tconcept of measurability and continuity have some important properties in common.

Definition of measurability:

A collection $R$ of subsets of a set $X$ is said to be $\sigma$-algebra in $X$ if $R$ has the following properties.

(i) \[X \in R\] So entire space is in collection.
(ii) if \[A \in R\] then \[A^c \in R\].
(iii) If \[A=\cup_{i=1}^{\infty} A_n\] and if \[A_n \in R\] for \[n=1,2,\cdots\] then \[A \in R\].

Just to show parallels, author also throws in definition of Topological space. But, then he never shows examples of measurable spaces. Perhaps he expects the readers to be familiar with these spaces. Perhaps...

Next section deals with review of metric spaces and topological definition of continuity. Plus he adds definition of continuity as a local property. Terseness of language in this book is fantastic! Definition of local continuity goes as follows:" A mapping $f$ of $X$ into $Y$ is said to be continuous at a point $x_0 \in X$ if for every neighborhood $V$ of $f(x_0)$, there corresponds a neighborhood $W$ of $x_0$, such that $f(W) \subset V$".

 This proposition is clear enough (watch the beautiful, terse language!). Next section deals with comments on measurability which is totally out of place. This should have been presented after the definition of measurability as an exercise.