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Math4121 L33
Continue on Lebegue integration
Sequence of functions
Proposition 6.4
Let f_n be a sequence of measurable functions, then \sup_n f_n,\inf_n f_n, \limsup_n f_n, \liminf_n f_n are measurable.
Proof:
Consider the set \{x\in \mathbb{R}, \sup_n f_n\leq c\}.
This is the set of x such that f_n(x)\leq c for all n.
\bigcap_{n=1}^{\infty} \{x\in \mathbb{R}, f_n(x)\leq c\} \subset \{x\in \mathbb{R}, \sup_n f_n(x)\leq c\}, by the definition of least upper bound.
Since the set on the right is intersection of measurable sets, it is measurable.
Therefore, \sup_n f_n is measurable.
The proof for \inf_n f_n, \limsup_n f_n, \liminf_n f_n are similar.
Consider {x\in \mathbb{R}, \inf_n f_n\leq c}=\bigcap_{n=1}^{\infty} \{x\in \mathbb{R}, f_n(x)\geq c\}.
\limsup_n f_n(x)=\inf_n \sup_{k\geq n} f_k(x) is measurable by \sup_{k\geq n} f_k(x) is measurable.
\liminf_n f_n(x)=\sup_n \inf_{k\geq n} f_k(x) is measurable by \inf_{k\geq n} f_k(x) is measurable.
QED
Lemma of function of almost everywhere
If f is measurable function and f(x)=g(x) for almost every x (on a set which the complement has Lebesgue measure 0), then g is measurable.
Proof:
Let c\in \mathbb{R}, F_1=\{x\in \mathbb{R}, f(x)>c\}, F_2=\{x\in \mathbb{R}, g(x)>c\}.
Recall the symmetric difference F_1\triangle F_2=\{x\in \mathbb{R}, f(x)\neq g(x)\}. By the definition of g, F_1\triangle F_2 has a measure 0.
In particular, all subsets of the F_1\triangle F_2 are measurable.
Notice that F_2=(F_1\setminus F_2)\cup (F_1\setminus (F_1\setminus F_2)).
Since F_1\setminus F_2 is measurable and F_1 is measurable, then F_2 is measurable.
QED
Example of measurable functions:
- Continuous functions are measurable.
\{x:f(x)>c\}=\{x:f(x)\in (c,\infty)\}=f^{-1}(c,\infty) is open (by open mapping theorem, or the definition of continuity in topology).
- Riemann integrable functions are measurable.
Outer content of the discontinuity of the function is 0.
\forall \sigma>0, where S_\sigma=\{x\in [a,b]: w(f,x)>\sigma\}, m(S_\sigma)=0.
S=\bigcup_{n=1}^{\infty} S_{\frac{1}{n}} has a measure 0. So f is continuous outside a set of measure 0.
m(S)\leq \sum_{n=1}^{\infty} m(S_{\frac{1}{n}})=0. So (detailed proof in the textbook)f agrees with a continuous function outside a set of measure 0. (almost everywhere)
Theorem 6.6
Let f_n be a sequence of measurable functions and f is a function satisfying \lim_{n\to\infty} f_n(x)=f(x) for almost every x (holds for sets which the complement has Lebesgue measure 0).
Then f(x)=\lim_{n\to\infty} f_n(x) is a measurable function.
Notice that f(x) is defined "everywhere"
Proof:
Apply the lemma of function of almost everywhere to the sequence f_n.
QED
Definition of simple function
A measurable function \phi:\mathbb{R}\to\mathbb{R} is called a simple function if it takes only finitely many values.
\text{range}(\phi)=\{d(x):x\in \mathbb{R}\}\subset \mathbb{R}
has finitely many values.
Equivalently, \exists \{a_1,a_2,\cdots,a_n\}\subset \mathbb{R} and disjoint measurable sets S_1,S_2,\cdots,S_n such that
\phi(x)=\sum_{i=1}^{n} a_i \chi_{S_i}(x)
where \chi_{S_i} is the indicator function of S_i.
Theorem 6.7
A function f is measurable if and only if there exists a sequence of simple functions \{\phi_n\} such that \lim_{n\to\infty} \phi_n(x)=f(x) for almost every x.
f is a limit of almost everywhere convergent sequence of simple functions.
(already proved backward direction)
Continue on Monday.