Math4121 Lecture 26

Lebesgue Measure

Lebesgue’s Integration

Partition on the y-axis, let $l$ be the minimum of $f(x)$ on the $y$ -axis, $L$ be the maximum of $f(x)$ on the $y$ -axis.

$l=l_0<l_1<\cdots<l_n=L$

Define $S_i=\{x\in[a,b]:l_{i-1}\leq f(x)<l_i\}$

Defined the characteristic function of set $S$ is $\chi_S(x)=1$ if $x\in S$ and $0$ otherwise.

Then $f$ lies between the following simple functions:

\sum_{i=1}^n l_{i-1}\chi_{S_i}\leq f(x)\leq \sum_{i=1}^n l_i\chi_{S_i}

This representation allows us to measure some weird sets on the $x$ -axis by constraining the $y$ -axis.

This is still kind of Riemann sum, but $S_i$ can be very weird (not just intervals).

If we can “measure” each $S_i$ , then we could define the integral of $f$ by

\sup_{l_0,\cdots,l_n}\sum_{i=1}^n l_{i-1}m(S_i)\quad \inf_{l_0,\cdots,l_n}\sum_{i=1}^n l_i m(S_i)

If we used Jordan content, for $m$ here, this is just a different perspective of Riemann integral.

If we use Borel measure maybe things would be different (perhaps, better)?

As we discussed last time, this limits the measurable sets significantly.

Nonetheless, let’s try

Characteristic function of the rational numbers

f(x)=\begin{cases} 1 & x\in\mathbb{Q}\cap[0,1]\\ 0 & \text{otherwise} \end{cases}

Take partition $0=x_0<x_1<\cdots<x_n=1$

$S_1=\{x:0\leq f(x)<\epsilon\}=[0,1]\setminus\mathbb{Q}$

$S_2=\{x:\epsilon\leq f(x)<1\}=\emptyset$

$S_3=\{x:1\leq f(x)<1+\epsilon\}=\mathbb{Q}\cap[0,1]$

\sum_{i=1}^n l_{i-1}m(S_i)=0\cdot 1+\epsilon\cdot 0+1\cdot 0=0

\sum_{i=1}^n l_i m(S_i)=\epsilon\cdot 1+1\cdot 0+(1+\epsilon)\cdot 0=\epsilon

So, $0\leq\int_0^1 f(x)dm\leq\epsilon$ , here $m$ means the measure we used.

As $\epsilon$ is arbitrary, we have $\int_0^1 f(x)dm=0$ .

This shows that $\int \chi_S(x)dm=m(S)$ for any measurable set $S$ .

Lebesgue’s Measure

Definition of Lebesgue measure

Outer Measure:

Given $S\subset [a,b]$ , let $\mathcal{C}$ be the collection of all countable covers of $S$ by open intervals.

m_e(S)=\inf_{C\in\mathcal{C}}m(C)

where $m(C)$ is the Borel measure.

Recall such $C$ are Boreal measurable because open interval are in $\mathcal{B}$ and $\mathcal{B}$ (being a sigma algebra) is closed under countable unions.

Properties:

Translation invariant: $m_e(S+a)=m_e(S)$
Countable additivity: $m_e(S\cup T)=m_e(S)+m_e(T)$ if $S\cap T=\emptyset$
$m([0,1])=1$

Notice we don’t have the difference property here.