2.5 KiB
Math4121 Lecture 30
Lebesgue Measure
\mathfrak{M}=\{S\subseteq\mathbb{R}:S\text{ is Lebesgue measurable}\} is a $\sigma$-algebra on \mathbb{R} (closed under complementation and countable unions).
Consequence of Lebesgue Measure
Every open set and closed set is Lebesgue measurable.
Inner and Outer Regularity of Lebesgue Measure
Outer regularity:
m_e(S)=\inf_{U\text{ open},S\subseteq U}m(U)
Inner regularity:
m_i(S)=\sup_{K\text{ closed},K\subseteq S}m(K)
Proof
Inner regularity:
Since m_i(S)=m(I)-m_e(I\setminus S), S\subseteq I for some closed interval I. Let \epsilon>0 and U be an open set such that I\setminus S\subseteq U and m(U)<m(I\setminus S)+\epsilon.
Take K=I\setminus U. Then K\subseteq S and K is closed and
m(K)=m(I)-m(U)>m(I)-m(I\setminus S)-\epsilon
So m_i(S)<m(K)+\epsilon. Since \epsilon is arbitrary, m_i(S)\leq m_e(S).
We can approximate m(S) from outside by open sets. If we are just concerned with "approximating" m(S), we can use finite union of intervals.
Symmetric difference
The symmetric difference of two sets S and T is defined as
S\Delta T=(S\setminus T)\cup(T\setminus S)
The XOR operation on two sets.
Theorem
If S\subset I is measurable, then for every \epsilon>0, \exists I_1,I_2,\cdots,I_n\subset I open intervals such that
m(S\Delta U)<\epsilon
where U=\bigcup_{j =1}^n I_j.
Proof
Let \epsilon>0 and m(V)<m(S)+\frac{\epsilon}{2}. Let K\subseteq S be closed set such that m(S)-\frac{\epsilon}{2}<m(K). V is an open cover of closed and bounded set K. By Heine-Borel theorem, K has a finite subcover. Let I_1,I_2,\cdots,I_n be the open intervals in the subcover.
Check:
m(S\Delta U)=m(S\setminus U)+m(U\setminus S)\leq m(S\setminus K)+m(U\setminus S)<\epsilon
Recall \{T_j\}_{j=1}^\infty are disjoint measurable sets. Then T=\bigcup_{j=1}^\infty T_j is measurable and
m(T)=\sum_{j=1}^\infty m(T_j)
Corollary (Better osgood's theorem on Lebesgue measure)
If S_1\subseteq S_2\subseteq S_3\subseteq\cdots are measruable (no need to be closed and bounded) and S=\bigcup_{j=1}^\infty S_j, then
m(S)=\lim_{j\to\infty}m(S_j)
Proof:
Let T_1=S_1 and T_i=S_i\setminus S_{i-1} for i\geq 2. Still have S=\bigcup_{j=1}^\infty T_j.
Where T_i are disjoint measurable sets. So m(S)=\sum_{j=1}^\infty m(T_j).
So \lim_{j\to\infty}m(S_j)=\sum_{j=1}^\infty m(T_j)=m(S).