4.2 KiB
Math 4121 Lecture 36
Random visit for Lebesgue Integration
Convergence Theorem
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:
To show the limit exists almost everywhere, let \epsilon>0, set E_n=\{x\in E: f_n(x)>\frac{A}{\epsilon}\}. We will show U=\bigcup_{n=1}^{\infty} E_n has measure <\epsilon. f_n(x)\geq \frac{A}{\epsilon}\chi_{E_n}(x), so
\frac{A}{\epsilon}m(E_n)=\int_E \frac{A}{\epsilon}\chi_{E_n}dm\leq \int_E f_n dm\leq A
In particular, m(E_n)<\epsilon. Since E_n\subset E_{n+1} for all n, m(U)=\lim_{n\to\infty} m(E_n)<\epsilon.
\lim_{n\to\infty} \int_E f_n dm\leq \int_E f dm\leq \lim_{n\to\infty}. To show the reverse inequality, let \phi be a simple function \leq f of the form \phi=\sum_{i=1}^{k} a_i\chi_{S_i} where S_i is sidjoint and \bigcup_{i=1}^{k} S_i\subseteq E.
Let \alpha\in (0,1) and set A_n=\{x\in S:f_n(x)-\alpha\phi(x)>0\}. This ensures that f_n(x)\geq \alpha\phi(x) for all x\in A_n.
Notice that A_n\subset A_{n+1} for all n and U=\bigcup_{n=1}^{\infty} A_n. lim_{n\to\infty} m(A_n\cap S_i)=m(S_i) for all i.
\int_{A_n} \phi dm=\sum_{i=1}^{k} a_i m(S_i\cap A_n).
As n\to\infty, m(A_n\cap S_i)\to m(S_i) and \sum_{i=1}^{k} a_i m(S_i\cap A_n)\to \int_E \phi dm.
Let \epsilon>0. There exists n_0 large such that \int_{A_n} \phi dm>\int_E \phi dm-\epsilon for all n\geq n_0.
Then for such n\geq n_0,
\int_E f_n dm\geq \int_{A_n} f_n dm\geq \int_{A_n} \alpha \phi dm>\alpha(\int_E \phi dm-\epsilon)
So, \lim_{n\to\infty} \int_E f_n dm\geq \alpha(\int_E \phi dm-\epsilon).
Since \alpha,\epsilon are arbitrary, set \alpha\to 1 and \epsilon\to 0 to get \lim_{n\to\infty} \int_E f_n dm\geq \int_E \phi dm.
For any simple function \phi\leq f, taking sup over all simple functions \phi\leq f gives \lim_{n\to\infty} \int_E f_n dm\geq \int_E f dm.
QED
Lemma Absolute Integrability
f is integrable on E if and only if |f| is integrable on E and \left|\int_E f \, dm\right|\leq \int_E |f| \, dm.
Proof:
If f^+ and f^- are integrable and |f|=f^+-f^-. So setting E_1=\{x\in E: f(x)\geq 0\} and E_2=\{x\in E: f(x)<0\}, these are disjoint and E=E_1\cup E_2.
\int_E |f| dm=\int_{E_1} f^+ dm+\int_{E_2} f^- dm
For the reverse inequality, note that
\int_E f^+ dm\leq \int_E |f| dm
and
\int_E f^- dm\leq \int_E |f| dm
QED
Corollary Properties of Integrals
Let f and g be integrable on E, and c\in \mathbb{R}.
\int_E (cf) dm=c\int_E f dm\int_E (f+g) dm=\int_E f dm+\int_E g dm
Proof:
First we prove it for f,g nonnegative and c\geq 0.
Take simple functions \phi_n\to f and \psi_n\to g pointwise. Then c\phi_n\to cf and \phi_n+\psi_n\to f+g pointwise.
By Monotone Convergence Theorem,
\int_E cf dm=\lim_{n\to\infty} \int_E c\phi_n dm=c\lim_{n\to\infty} \int_E \phi_n dm=c\int_E f dm
Second part leave as homework.
QED
Theorem 6.8
Riemann integrable functions are Lebesgue integrable and the values of the integrals are the same.
Proof:
Say f is Riemann integrable on [a,b]. m\leq f(x)\leq M for all x\in [a,b].
We can find a partition P_n\subseteq P_{n+1} of [a,b] such that L(P_n,f)\nearrow \int_a^b f dx and U(P_n,f)\searrow \int_a^b f dx.
Let \phi_n=\sum_{i=1}^{k} m_i \chi_{I_i} and \psi_n=\sum_{i=1}^{k} M_i \chi_{I_i} where I_i is an interval in P_n.
So \int_a^b \phi_n dm=L(P_n,f) and \int_a^b \psi_n dm=U(P_n,f).
m\leq \phi_n\leq f\leq \psi_n\leq M for all n. almost everywhere.
By Monotone Convergence Theorem, to \phi_{n-m} we have g(x)=\lim_{n\to\infty} \phi_n(x), h(x)=\lim_{n\to\infty} \psi_n(x) exists for almost every x\in [a,b].
g(x)\leq f(x)\leq h(x) almost everywhere.
So
\int_a^b g dm= \int_a^b f dm= \int_a^b h dm =\int_a^b f dx
So h(x)=f(x)=g(x) almost everywhere.
QED