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--- math105-s22:notes:lecture_5 [2022/01/31 23:32]
pzhou [Pugh 6.2: construction of measure]
+++ math105-s22:notes:lecture_5 [2022/02/01 23:35] (current)
pzhou
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 ====== Lecture 5 ======
 $\gdef\mcal{\mathcal{M}}$
+[[https://berkeley.zoom.us/rec/share/xjmorYXsD5aEU3_mS6oHTOc413MzXAOIKlj5v1LUWOksJVeNeLzhp-QSVBff_AEF.sRwBxAiQgwZnO3D9 | video ]]
 Welcome back to in-person instruction. I will continue type in here as a way to prepare for class.
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-One statement worth emphasizing is that, "adding or removing a null-set does not affect measurability". If $Z$ is a null-set, then for any subset $A$, we have $\omega(A) \leq \omega(A \cup Z) \leq \omega(A) + \omega(Z) = \omega(A)$, hence $\omega(A \cup Z) = \omega(A)$. Similarly, $\omega(A \cap Z^c) = \omega( (A \cap Z^c) \cup (Z \cap A) ) = \omega(A)$, note $Z \cap A$ is null as well.
+One statement worth emphasizing is that, "adding or removing a null-set does not affect measurability". If $Z$ is a null-set, then for any subset $A$, we have $\omega(A) \leq \omega(A \cup Z) \leq \omega(A) + \omega(Z) = \omega(A)$, hence $\omega(A \cup Z) = \omega(A)$. Similarly, $\omega(A \cap Z^c) = \omega( (A \cap Z^c) \cup (Z \cap A) ) = \omega(A)$, note $Z \cap A$ is null as well. Thus, adding or removing $Z$ does not affect the outer-measure. Hence, does not affect the measurability of $E$.
+Hyperplanes $\{a\}\times \R^{n-1} \In \R^n$ is a null-set. For example, for any $\epsilon>0$, we can cover $\{0\} \times \R$ by
+$$ \{0\} \times \R = \cup_{n=1}^\infty (-\epsilon 2^{-2n-2}, \epsilon 2^{-2n-2}) \times (-2^n, 2^n) $$
+where the sum of area of boxes is less than $\epsilon$.
+===== Pugh 6.4: Regularity =====
+Our goal here is to prove that, any Lebesgue measurable set is a Borel set plus or minus a null-set. More precisely. $E$ is Lebesgue measurable, if and only if there is a $G_\delta$-set (countable intersection of open) $G$, and an $F_\sigma$-set, $F$, where $F \In E \In G$, such that $m(G \RM F) = 0$ (why not asking $m(G) = m(F)$? )

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