NoteNextra-origin/content/Math4121/Math4121_L34.md

Math4121 Lecture 34

Important:

\mathfrak{M}=\{S\subset \mathbb{R}: S \text{ satisfies the caratheodory condition}\}, that is, for any X of finite outer measure,

m_e(X)=m_e(X\cap S)+m_e(X\cap S^c)

In particular, the measure of sets can be infinite, not necessarily bounded. (We want to make the real line measurable.)

Lebesgue Integral

Simple Function

A function \phi is called a simple function if

\phi(x)=\sum_{i=1}^{n} a_i \chi_{S_i}(x)

where a_i\in \mathbb{R} and \chi_{S_i}=\begin{cases}1, & x\in S_i \\ 0, & x\notin S_i\end{cases}

where \{S_i\}_{i=1}^{n} are pairwise disjoint each having finite measure.

constant function is not simple (\mathbb{R} is not finite measurable sets.)

Theorem 6.6

A function f is measurable on [a,b] if and only if there exists a sequence of simple functions \{\phi_n\} such that \lim_{n\to\infty} \phi_n(x)=f(x) almost everywhere on [a,b].

Proof:

Partition [-n,n] into n2^{n+1} pieces.

(These are just horizontal strips from -n to n with width \frac{1}{2^n}.)

E_{n,k}=\{x\in[-n,n]:\frac{k}{2^n}\leq f(x)<\frac{k+1}{2^n}\}

for -n2^n<k<n2^n

E_{n,n2^n}=\{x\in[-n,n]:f(x)\geq n\}

E_{n,-n2^n}=\{x\in[-n,n]:f(x)<\frac{-n2^n+1}{2^n}\}

\phi_n(x)=\frac{k}{2^n}\chi_{E_{n,k}}(x)

is a simple function.

We need to justify that \phi_n(x)\to f(x) for all x\in\mathbb{R}.

Let x\in\mathbb{R}. And choose n_0 large such that x\in [-n_0,n_0] and f(x)\in [-n_0,n_0].

Then, for n\geq n_0,

|\phi_n(x)-f(x)|<\frac{1}{2^n}\to 0

as n\to\infty.

QED

Integration

Given a measurable set E and a simple function \phi, we define

\int_E \phi dm=\sum_{i=1}^{n} a_i m(E\cap S_i)

Properties 6.10

Let \phi and \psi be simple functions, c\in \mathbb{R}, and E=E_1\cup E_2 where E_1\cap E_2=\emptyset and E_1,E_2\in \mathfrak{M}. Then,

1. \int_E c\phi dm=c\int_E \phi dm (linearity)
2. \int_E (\phi+\psi)dm=\int_E \phi dm+\int_E \psi dm (additivity of simple functions)
3. if \phi(x)\leq \psi(x) for all x\in E, then \int_E \phi dm\leq \int_E \psi dm (monotonicity)
4. \int_E \phi(x)dm=\int_{E_1} \phi(x)dm+\int_{E_2} \phi(x)dm (additivity over disjoint measurable sets)

Proof:

Let \phi(x)=\sum_{i=1}^{n} a_i \chi_{S_i}(x) and \psi(x)=\sum_{j=1}^{m} b_j \chi_{T_j}(x).

\phi+\psi=\sum_{i=1}^{n} a_i \chi_{S_i}+\sum_{j=1}^{m} b_j \chi_{T_j}

Without loss of generality, we may assume that x\in E, \bigcup_{i=1}^{n} S_i=\bigcup_{j=1}^{m} T_j=E.

So

\phi+\psi=\sum_{i,j=1}^{n,m}(a_i+b_j) \chi_{S_i\cup T_j}

is a simple function.

\begin{aligned}
\int_E (\phi+\psi)dm&=\sum_{i,j=1}^{n,m}(a_i+b_j) m(E\cap S_i\cup T_j) \\
&=\sum_{i=1}^{n} a_i \sum_{j=1}^{m} m(E\cap S_i\cup T_j)+\sum_{j=1}^{m} b_j \sum_{i=1}^{n} m(E\cap S_i\cup T_j) \\
&=\sum_{i=1}^{n} a_i m(E\cap S_i)+\sum_{j=1}^{m} b_j m(E\cap T_j) \\
&=\int_E \phi dm+\int_E \psi dm
\end{aligned}

\phi(x)=\sum_{i=1}^{n} a_i\sum_{j=1}^{m} \chi_{S_i\cap T_j}(x)

\psi(x)=\sum_{i=1}^{n} b_i\sum_{j=1}^{m} \chi_{S_i\cap T_j}(x)

If x\in S_i\cap T_j, then \phi(x)=a_i and \psi(x)=b_j, therefore a_i\leq b_j.

So,

\begin{aligned}
\int_E \phi dm&=\sum_{i=1}^{n} \sum_{j=1}^{m} a_i m(E\cap S_i\cap T_j) \\
&\leq \sum_{i=1}^{n} \sum_{j=1}^{m} b_i m(E\cap S_i\cap T_j) \\
&=\int_E \psi dm
\end{aligned}

QED

Back on Wednesday.