NoteNextra-origin/content/Math4121/Math4121_L32.md

Math4121 Lecture 32

Chapter 6: The Lebesgue Integral

Measurable Functions

Definition: A function f:\mathbb{R}\to\mathbb{R} is measurable on the interval [a,b] if \{x\in [a,b]:f(x) > c\} is measurable for all c\in\mathbb{R}, called the super level set of f.

Denote \{f> c\}

Proposition 6.1

The following are equivalent:

For all c\in\mathbb{R},

1. \{x\in [a,b]:f(x) > c\} is measurable.
2. \{x\in [a,b]:f(x) < c\} is measurable.
3. \{x\in [a,b]:f(x) \geq c\} is measurable.
4. \{x\in [a,b]:f(x) \leq c\} is measurable.
5. \{x\in \mathbb{R}:c \leq f(x) < d\} is measurable for all c,d\in\mathbb{R}.

Proof:

Since the complement of a measurable set is measurable. (1) \iff (4). and (2) \iff (3).

We only need to show (1) \implies (2).

Since \{f>c\}=\bigcup_{n=1}^{\infty}\{f\geq c+\frac{1}{n}\}.

So (1) \implies (2).

Since (2) \implies (1)-(4) \implies (5).

To see (5) \implies (1), we have \{f\geq c\}=\bigcup_{n=1}^{\infty}\{x\in\mathbb{R}:c \leq f(x) < c+n\}

QED

Proposition 6.3

Let f,g be measurable on [a,b] and \alpha\in\mathbb{R}. Then the following are measurable:

1. f+g
2. fg
3. \alpha f
4. |f|^\alpha

Proof:

If \alpha=0, then \alpha f and |f|^\alpha are constant functions, hence measurable.

But for constant functions h, ${h>c}=\begin{cases} \emptyset & \text{if } c\geq h \ \mathbb{R} & \text{if } c < h \end{cases}$

For a\neq 0, \{x\in \mathbb{R}:\alpha f(x) > c\}=\{x\in \mathbb{R}:f(x) > \frac{c}{\alpha}\}

Similarly, \{x\in \mathbb{R}:|f(x)|^\alpha > c\}=\{x\in \mathbb{R}:|f(x)| > c^{1/\alpha}\}=\{x\in \mathbb{R}:f(x) > c^{1/\alpha}\}\cup\{x\in \mathbb{R}:f(x) < -c^{1/\alpha}\}

We want to show \{f+g>c\}=\bigcup_{q\in \mathbb{Q}}\{f>q\}\cap\{g>c-q\}

\{f+g>c\}\supseteq \bigcup_{q\in \mathbb{Q}}\{f>q\}\cap\{g>c-q\}

if x is in the RHS, then \exists q\in \mathbb{Q} such that f(x) > q and g(x) > c-q, therefore f(x)+g(x) > c.

So x is in the LHS.

\{f+g>c\}\subseteq \bigcup_{q\in \mathbb{Q}}\{f>q\}\cap\{g>c-q\}

Let x\in \mathbb{R} such that f(x)+g(x) > c. Need to find q\in \mathbb{Q} such that f(x) > q and g(x) > c-q.

Since f(x)>c-g(x), by the density of \mathbb{Q} in \mathbb{R}, \exists q\in \mathbb{Q} such that q > c-g(x).

For fg, we have fg=\frac{1}{4}((f+g)^2-(f-g)^2)

So fg is measurable.

QED

Limit of Measurable Functions

Proposition 6.4

Let \{f_n\} be a sequence of measurable functions on [a,b]. Then,

g(x)=\sup_{n\geq 1}f_n(x),\inf_{n\geq 1}f_n(x),\limsup_{n\to\infty}f_n(x),\liminf_{n\to\infty}f_n(x)

are all measurable functions

Corollary of Proposition 6.4

If \{f_n\}_{n=1}^{\infty} are measurable functions and f(x)=\lim_{n\to\infty}f_n(x) exists for all x\in[a,b], then f is measurable. (pointwise limit of measurable functions is measurable)

Definition of almost everywhere

A property holds almost everywhere if it holds everywhere except for a set of measure zero.