3.1 KiB
Math4121 Lecture 35
Continue on Lebesgue Integration
Lebesgue Integration
Definition of Lebesgue Integral
For simple functions \phi = \sum_{i=1}^{n} a_i \chi_{S_i}, given a measure E, the Lebesgue integral is defined as:
\int_{\mathbb{R}^n} \phi \, dm = \sum_{i=1}^{n} a_i m(S_i\cap E)
Given a non-negative measurable function f and a measurable set E.
Define \int_E f \, dm = \sup \left\{ \int_E \phi \, dm : \phi \text{ is a simple function and } \phi \leq f \right\}
(We do allows $\int_E f , dm = \infty$)
For general measurable function f, we can define f^-(x)=\max\{0,-f(x)\}, f^+(x)=\max\{0,f(x)\}. (The positive part of the function and the negative part of the function, both non-negative)
Then f=f^+-f^-.
We say f is integrable if \int_E f^+ \, dm < \infty and \int_E f^- \, dm < \infty. (both finite) If at least one is finite, define
\int_E f \, dm = \int_E f^+ \, dm - \int_E f^- \, dm
We allow for A-\infty = -\infty and A+\infty = \infty for any A\in \mathbb{R}. But not \infty-\infty.
Immediate Properties of Lebesgue Integral
If f is measurable and m(E)=0, then \int_E f \, dm = 0.
If E=E_1\cup E_2 and E_1\cap E_2=\emptyset, then \int_E f \, dm = \int_{E_1} f \, dm + \int_{E_2} f \, dm.
Corollary
If f\leq g almost everywhere, (f\leq g except for a set of measure 0), then \int_E f \, dm \leq \int_E g \, dm.
Proof:
Let F=\{x\in E: f(x)>g(x)\}. Then m(F)=0.
\begin{aligned}
\int_E f \, dm &= \int_{E\setminus F} f \, dm + \int_F f \, dm\\
&\leq \int_{E\setminus F} g \, dm+\int_E g \, dm\\
&= \int_E g \, dm
\end{aligned}
QED
Proposition 6.13
If f is non-negative and \int_E f \, dm =0, then f=0 almost everywhere on E, f(x)=0 \forall x\in E\setminus F, where m(F)=0.
Proof:
Let E_n=\{x\in E: f(x)\geq \frac{1}{n}\}. Then \frac{1}{n}\chi_{E_n}(x)\leq f(x) for all x\in E.
By definition \frac{1}{n}m(E_n)=\int_E \frac{1}{n}\chi_{E_n} \, dm \leq \int_E f \, dm =0.
Therefore, m(E_n)=0 for all n.
Now U=\{x\in E: f(x)>0\}=\bigcup_{n=1}^{\infty} E_n, and E_n\subseteq E_{n+1} for all n.
Therefore, m(U)=m(\bigcup_{n=1}^{\infty} E_n)=\lim_{n\to\infty} m(E_n)=0.
QED
Convergence Theorems
When does \lim_{n\to\infty} \int_E f_n \, dm = \int_E \lim_{n\to\infty} f_n \, dm?
Theorem 6.14 Monotone Convergence Theorem
Let \{f_n\} be a monotone increasing sequence of measurable functions on E and f_n\to f almost everywhere on E. (f_n(x)\leq f_{n+1}(x) for all x\in E and n)
If there exists A>0 such that \left|\int_E f_n \, dm\right|\leq A for all n\in \mathbb{N}, then f(x)=\lim_{n\to\infty} f_n(x) exists for almost every x\in E and it is integrable on E and
\int_E f \, dm = \lim_{n\to\infty} \int_E f_n \, dm
Proof:
First to show the limit exists almost everywhere. It suffices to show
\mathcal{U}=\{x\in E: f_n(x) \text{ is unbounded}\}
has measure 0.
Let \epsilon>0 and write
U=\bigcup_{n=1}^{\infty} E_n
where E_n=\{x\in E: |f_n(x)|\geq \epsilon\}.
Then U\subseteq \mathcal{U} and m(U)<\epsilon.
CONTINUE NEXT TIME.
QED