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)

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


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)

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