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 $f$ agrees with a continuous function outside a set of measure $0$. (almost everywhere) (detailed proof in the textbook)

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.

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

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.