Math4121 Lecture 28

$\implies$ If Lebesgue criterion holds for $S$, then for any $X$ of finite outer measure,

$$m_e(X)=m_e(X\cap S)+m_e(X\setminus S)$$

First, we extend Lebesgue criterion to intervals $I$ that may not contain $S$. Then we can find $J,K$ intervals neighboring $I$ such that $S\subseteq \tilde{I}=J\cup I\cup K$.

By Lebesgue criterion,

$$\begin{aligned} m_e(\tilde{I})&=m_e(\tilde{I}\cap S)+m_e(\tilde{I}\setminus S)\\ &=m_e(S)+m_e(\tilde{I}\setminus S)\\ &=m_e(S^c\cap \tilde{I})+m_e(S\cap \tilde{I})\\ &=\sum_{L\in \{J,I,K\}}m_e(L\cap S^c)+m_e(L\cap S)\\ &\geq \sum_{L\in \{J,I,K\}}m_e(L)\\ &=m_e(\tilde{I}) \end{aligned}$$

Therefore, $m_e(I)=m_e(S^c\cap I)+m_e(S\cap I)$.

Now, let $X$ has finite outer measure, let $\epsilon>0$, we can find $\{I_j\}_{j=1}^{\infty}$ covering $X$ and

$$\sum_{j=1}^{\infty} \ell(I_j)<m_e(X)+\epsilon$$ $$\begin{aligned} m_e(X)&\leq m_e(X\cap S)+m_e(S^c\cap X)\\ &\leq m_e\left(\bigcup_{j=1}^{\infty} I_j\cap S\right)+m_e\left(\bigcup_{j=1}^{\infty} I_j\cap S^c\right)\\ &\leq \sum_{j=1}^{\infty} m_e(I_j\cap S)+m_e(I_j\cap S^c)\\ &=\sum_{j=1}^{\infty} m_e(I_j)\\ &<m_e(X)+\epsilon \end{aligned}$$

First we prove $m_e(\bigcup_{j=1}^{\infty} S_j)=\sum_{j=1}^{\infty} m(S_j)$ by induction.

$n=1$ is trivial.

Let $n>1$ and suppose the statement holds for $n-1$. Take $X=\bigcup_{j=1}^{n-1} S_j$, then $S_n\cap X=S_n, X\setminus S_n=\bigcup_{j=1}^{n-1} (S_j)$.

By Caratheodory’s criteria,

$$\begin{aligned} m_e(X)&=m_e(S_n)+m_e(\bigcup_{j=1}^{n-1} S_j)\\ m_e(\bigcup_{j=1}^{n} S_j)&=m(S_n)+\sum_{j=1}^{n-1} m(S_j)\\ &=\sum_{j=1}^{n} m(S_j) \end{aligned}$$

Take the limit as $n\to\infty$, and justify this.

$\sum_{j=1}^{\infty} m(S_j)=m_e(\bigcup_{j=1}^{\infty} S_j)\leq m_e(S)$

Since $m_e(S)$ is finite and $m(S_j)$ is monotone, the limit exists.

Therefore, $\sum_{j=1}^{\infty} m(S_j)\leq m_e(S)\leq \sum_{j=1}^{\infty} m(S_j)$

So $S$ is measurable.

Let $S_1, S_2$ be measurable sets.

We prove by verifying the Caratheodory’s criteria for $S_1\cup S_2$.