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