$\gdef\mcal{\mathcal{M}}$

video

Welcome back to in-person instruction. I will continue type in here as a way to prepare for class.

After a long toil of last two weeks, we have established the existence of measurable sets and Lebesgue measure on $\R^n$. We know open sets and closed sets are measurable, and countable operations won't take us away from measurable sets. The Lebesgue measure on measurable sets satisfies all the intuitive properties that you wish it has.

After an actual reading of Tao's 7.5, I decided that it is a bit misleading (especially the part that composition of measurable functions are not automatic measurable (but will turns out to be so after some work). Let's first review what should be true in general.

Let $S$ be a set, and $2^S$ be the set of subsets in $M$.

$\sigma$-algebra : Let $\mcal_S$ be a subset of $2^S$. We say $\mcal_S$ is a $\sigma$-algebra, if it contains $\emptyset$ and is closed under taking complement and countable intersections. (Hence it contains $S$ itself, and is closed under countable union)

We refer to the pair of a space and a $\sigma$-algebra, $(S, \mcal_S)$, a measurable space.

measure on $(S, \mcal_S)$ : A measure is a function $\omega: \mcal_S \to [0, \infty]$, such that $\omega(\emptyset) = 0$ and satisfies countable additivity.

The triple $(S, \mcal_S, \omega)$ is called a measure space .

measurable function , let $(X, \mcal_X)$ and $(Y, \mcal_Y)$ be two measurable spaces, a function $f: X \to Y$ is measurable if for any $V \in \mcal_Y$, we have $f^{-1}(V) \in \mcal_X$.

This may reminds you of the definition of topological spaces and continuous maps.

Hence by definition, composition of measurable sets are measurable. Why we care about composition of measurable set? It is useful in ergodic theory, which is about iterations of $f^n$, where $f: X \to X$ is measurable.

If $S$ happens to also be a topological space, with $\tau_S$ denote the set of open sets, we may define Borel $\sigma$-algebra $\mathcal{B}_S$ , which is the smallest $\sigma$-algebra of $2^S$ that contains $\tau_S$. If a subset of $S$ is in the Borel sigma-algebra, we call it a Borel set. In particular, countable intersection of open sets ($G_\delta$-set) and countable union of closed set ($F_\sigma$-set) are Borel set.

Now, you may ask, consider $\R$ with the usual topology, are Borel sets equivalent with Lebesgue measurable set? Not quite. They may differ by a measure zero subset (called null set, or zero set).

Let's turn to Pugh now. We first need to revisit Theorem 6.5, page 389. Let $S$ be any set. One can construct a $\sigma$-algebra and a measure on it, starting from any outer-measure $\omega: 2^S \to [0, \infty]$. An outer-measure $\omega$ on $S$ is any such function that satisfies $\omega(\emptyset)=0$, monotonicity, and sub-additivity.

From outer-measure $\omega$, one can define measurable set on $S$, using Caratheory criterion. Namely, $E$ is measurable if and only if for any set $A$, we have $\omega(A) = \omega(A \cap E)+ \omega(A \cap E^c)$. We need to show that, measurable set forms a $\sigma$-algebra. The proof is no different than Tao 7.4.8. In short, Pugh's theorem 5 is a 'free upgrade' of Tao's result, the proof of Tao goes through verbatim.

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 $\epsilon$.

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 $m(G) = m(F)$? )