Math4121 Lecture 29

Continue on Measure Theory

Lebesgue Measure

Caratheodory’s criterion:

$S$ is Lebesgue measurable if for all $A\subset S$ ,

m_e(X) = m_e(X\cap S) + m_e(X\cap S^c)

Let $\mathfrak{M}$ be the collection of all Lebesgue measurable sets.

$\phi\in\mathfrak{M}$
$\mathfrak{M}$ is closed under countable unions (proved last lecture)
$\mathfrak{M}$ is closed under complementation ( $\mathfrak{M}$ is a $\sigma$ -algebra) (goal today)

Desired properties of a measure:

$m(I)=\ell(I)$ for all intervals $I$

If $\{S_n\}_{n=1}^{\infty}$ is a set of pairwise disjoint Lebesgue measurable sets, then

$m\left(\bigcup_{n=1}^{\infty}S_n\right) = \sum_{n=1}^{\infty}m(S_n)$ 3. If $R\subset S$ , then $m(S\setminus R) = m(S) - m(R)$

Recall the Borel $\sigma$ -algebra $\mathcal{B}$ was the smallest $\sigma$ -algebra containing closed intervals. Therefore $\mathcal{B}\subset\mathfrak{M}$ .

Towards proving $\mathfrak{M}$ is closed under countable unions:

Theorem 5.9 (Finite union/intersection of Lebesgue measurable sets is Lebesgue measurable)

Any finite union/intersection of Lebesgue measurable sets is Lebesgue measurable.

Proof

Suppose $S_1, S_2$ is a measurable, and we need to show that $S_1\cup S_2$ is measurable. Given $X$ , need to show that

Finite union cut

m_e(X) = m_e(X_1\cup X_2\cup X_3)+ m_e(X_4)

Since $S_1$ measurable, $m_e(X_1\cup X_2\cup X_3)=m_e(X_3)+m_e(X_1\cup X_2)$ .

Since $S_2$ measurable, $m_e(X_3\cup X_4)=m_e(X_3)+m_e(X_4)$ .

Therefore,

\begin{aligned} m_e(X) &= m_e(X_1\cup X_2\cup X_3) + m_e(X_4) \\ &= m_e(X_1\cup X_2) + m_e(X_3)+m_e(X_4) \\ &= m_e(X_1\cup X_2) + m_e(X_3\cup X_4) \\ &= m_e(X) \end{aligned}

by measurability of $S_1$ again.

Theorem 5.10 (Countable union/intersection of Lebesgue measurable sets is Lebesgue measurable)

Any countable union/intersection of Lebesgue measurable sets is Lebesgue measurable.

Proof

Let $\{S_j\}_{j=1}^{\infty}\subset\mathfrak{M}$ . Definte $T_j=\bigcup_{k=1}^{j}S_k$ such that $T_{j-1}\subset T_j$ for all $j$ .

And $U_1=T_1$ , $U_j=T_j\setminus T_{j-1}$ for $j\geq 2$ .

Then $\bigcup_{j=1}^{\infty}S_j=\bigcup_{j=1}^{\infty}T_j=\bigcup_{j=1}^{\infty}U_j$ . Notice that $\{U_j\}_{j=1}^{\infty}$ are pairwise disjoint, and $\{T_j\}_{j=1}^{\infty}$ are monotone.

Let $X$ have finite outer measure. Since $U_n$ is measurable,

\begin{aligned} m_e(X\cap T_n) &= m_e(X\cap T_n\cap U_n)+ m_e(X\cap T_n\cap U_n^c) \\ &= m_e(X\cap U_n)+ m_e(X\cap T_{n-1}) \\ &= \sum_{j=1}^{n}m_e(X\cap U_j) \end{aligned}

Since $T_n$ is measurable and $T_n\subset S$ , $S^c\subset T_n^c$ . $m_e(X\cap T_n^c)\geq m_e(X\cap S^c)$ .

Therefore,

m_e(X)=m_e(X\cap T_n)+m_e(X\cap T_n^c)\\ \geq \sum_{j=1}^{n}m_e(X\cap U_j)+m_e(X\cap S^c)

Take the limit as $n\to\infty$ ,

\begin{aligned} m_e(X) &\geq \sum_{j=1}^{\infty}m_e(X\cap U_j)+m_e(X\cap S^c) \\ &= m_e(\bigcup_{j=1}^{\infty}(X\cap U_j))+m_e(X\cap S^c) \\ &= m_e(X\cap S)+m_e(X\cap S^c) \\ &\geq m_e(X) \end{aligned}

Therefore, $m_e(X\cap S)=m_e(X)$ .

Therefore, $S$ is measurable.

Corollary from the proof

Every open or closed set is Lebesgue measurable.

(Every open set is a countable union of disjoint open intervals)