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We finished Pugh 6.6. We mainly go over $\int f+g = \int f + \int g$.

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We will first follow Pugh's approach, then we will cover Tao's approach in exercises.

• Use undergraph of a non-negative function to define measurability and its measure. If the measure is finite, then call this function integrable.
• Monotone convergence theorem. (Recall upward/downward continuity theorem)
• Completed undergraph (not the closure of the undergraph, but just fiberwise closure). Can be used interchangeably with the undergraph
• upper and lower envelope sequence of a function, just like how one define the liminf and limsup.
• Dominated Convergence theorem.
• Many examples: running bump, shrinking bumps.

Discussion question:

• In Tao, one define measurable function $f: \R \to \R$, to be such that pre-image of open sets are measurable. Does this agree with Pugh's definition using undergraph?
• Pugh Ex 25, 28

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If $E \In \R^n, F \In \R^k$ are measurable, then $E \times F$ is measurable, with $m(E) \times m(F) = m(E \times F)$.

Let's first treat some special case. If $m(E)=0$, and $m(F) = \infty$, what is $m(E \times F)$? You have seen a special case as $m ( \{ 0 \} \times \R)=0$ in $\R^2$. The general proof is similar, for each $\epsilon$, and each $n \in \N$, we can find a countable collection of boxes that covers $E \times B(0, n)$ with total volume less than $\epsilon/2^n$. Then, we let $n=1,2,\cdots$, and put together these collection of boxes into a bigger collection (still countable), that gives a cover of $E \times \R^k$ with total area less than $\epsilon$.

Next, let's prove some nice cases, that $m(E \times F) = m(E) m(F)$.

• $E, F$ are open boxes.
• $E, F$ are open sets (each open set is a disjoint union of open boxes and a measure zero set. See Pugh's dyadic proof).

Now, how to prove that $ E \times F$ is measurable? We could use Caratheodory criterion, or, we could use Lebesgue criterion, by constructing outer and inner approximations. Again, we may assume $E$ is bounded and $F$ is bounded (otherwise, they can be written as disjoint union of bounded measurable pieces, and we can deal with them pieces by pieces, and do countable union in the end). We may assume $E \In B_1, F \In B_2$, for $B_i$ some open boxes.

Let $H_E$ be a $G_\delta$-set and $K_E$ be $F_\delta$, such that $H_E \supset E \supset K_E$, and $m(H_E \RM K_E) = 0$. Define $H_F, K_F$ for $F$ similarly. Then, by downward monotone continuity of measure, we have $$ m(H_E \times H_F) = \lim m(H_{E,n} \times H_{F,n}) = \lim m(H_{E,n}) \times m(H_{F,n}) = m(H_E) \times m(H_F) = m(E) \times m(F) $$ where $H_{E,n}$ are open sets, with $H_{E,n} \supset H_{E,n+1}$, and $H_E = \cap_n H_{E,n}$ for all $n$.

Also, we have $$ H_E \times H_F \RM (K_E \times K_F) \subset (H_E \times K_E) \times B_2 \cup B_1 \times (H_F \times K_F) $$ where the last term is a null set, hence $m(K_E \times K_F) = m(H_E \times H_F) = m(E) \times m(F)$.

If $E \In \R^n \times \R^k$ is measurable, then $E$ is a zero set if and only if almost( = up to a zero set) every slice $E_x$, ($x \in \R^n$) is measure zero.

Pf: as usual, we may assume $n=k=1$ and $E$ is contained in the unit square. Suppose $E$ is measurable, and $m(E_x)=0$ for almost all $x \in I$, then we want to show $m(E)=0$. Let $Z \In I$ be the set of $x$ where $m(E_x) \neq 0$. Then, $m(Z)=0$. Since $Z \times I$ is measureable and has measure 0, we may replace $E$ by $E \RM (Z \times I)$, and assume for all $x \in I$, $m(E_x)=0$.

By inner regularity, we may replace $E$ by a closed set $K$. Since E is bounded, hence $K$ is compact. Now, we try to cover $K$ by open boxes of total area less than $\epsilon$. Let $K_1= \pi (K)$ the projection to the first factor, than $K_1$ is compact.

• For each $x \in I$, we cover $K_x$ by an open set $V(x)$ of $m(V(x))<\epsilon$. We can find $U(x) \supset x$, that $U(x) \times V(x) \supset \pi_1^{-1}(U(x)) $. This is possible since $K$ is compact.
• We know $K \subset \cup_x U(x) \times V(x)$, but that's uncountably many set. We can pass to a finite subcover, indexed by $x_1, \cdots, x_N$. Let $U_i = U(x_i) \RM (\cup_{j<i} U(x_j))$, $V_i = V(x_i)$, then we still have $U_i \times V_i \supset \pi^{-1} U_i$. Thus, $U_i$ are disjoint, and we have $m (\cup U_i) \leq 1$ and $m (K) \leq \sum_i m(U_i)\times m(V_i) \leq \epsilon$.

1. Can you prove that $\{y=x\} \In \R^2$ has measure $0$?
2. In both of the two proofs above, we assumed $E$ was bounded, how to deal with the general case?
3. Prove that every closed subset (e.g. your favorite Cantor set is a closed set) in $\R$ is a $G_\delta$-set. Is it true that every open set is a $F_\sigma$-set?

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

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

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If $A \In B$ are both measurable, then $B \RM A$ is measurable, and $m(B \RM A) = m(B) - m(A)$.

We need to show that for any subset $E \In \R^n$… wait a second, do we really need to go by definitions again? After all these preparations, we should be able to exploit our sweat. Hint

• $B \RM A = B \cap A^c$.
• $B = A \sqcup (B \RM A)$, a decomposition into measurable subsets.

Let $\{E_j\}_{j=1}^\infty$ be a countable collection of disjoint subsets. Then, their union $E$ is measurable, and we have

$$ m^*(E) = \sum_{j=1}^\infty m^*(E_j) $$

Proof: Tao's proof is really clever, let's first try to go through the proof, then discuss how we can come up with it ourselves.

First, we start by showing $E$ is measurable from the definition: we want to show for any subset $A$, we have $$ m^*(A) = m^*(A \cap E) + m^*(A \RM E) $$ Suffice to show $\geq$ direction. Let $F_N = \sum_{j=1}^N E_j$. We have two expressions

• $m^*(A \cap E) \leq \sum_{j=1}^\infty m(A \cap E_j) = \sup_{N > 1} \sum_{j=1}^N m^*(A \cap E_j) = \sup_{N > 1} m^*(A \cap F_N) $
• $m^*(A \RM E) \leq m^*(A \RM F_N)$ for all $N$

Hence, $$ m^*(A \RM E) + m^*(A \cap E) \leq \sup_{N > 1} (m^*(A \cap F_N)) + m^*(A \RM E)) = \sup_{N > 1} (m^*(A \cap F_N)) + m^*(A \RM F_N)) = \sup_{N > 1} m^*(A) = m^*(A) $$

OK, that shows $E$ is measurable. To finish off, we need to show countable addivity $$ m^*(E) = \sum_{j=1}^\infty m^*(E_j) $$ Since $E = \cup E_j$, we have $\leq$ from countable sub-addivity. Then, by monotonicity, we have $$ m^*(E) \geq m^*(F_N) = \sum_{j=1}^N m^*(E_j) $$ since this is true for all $N$, we can sup over $N$, and get $$ m^*(E) \geq \sup_N \sum_{j=1}^N m^*(E_j) = \sum_{j=1}^\infty m^*(E_j) $$

One slogan is to approximate $E$ by $F_N$. We want to prove $$ m^*(A) \geq m^*(A \cap E) + m^*(A \RM E) $$ If we have

• (1) $m^*(A \cap E) \leq m^*(A \cap F_N)$ and
• (2) $m^*(A \RM E) \leq m^*(A \RM F_N)$,

then we can write $$ m^*(A \cap E) + m^*(A \RM E) \leq m^*(A \cap F_N) + m^*(A \RM F_N) = m^*(A) $$ but unfortunately, (1) is wrong. One way to remedy this, is to show that (assuming $m^*(A)< \infty$), for any $\epsilon > 0$, there exists an $N$, such that $m^*(A \cap E) \leq m^*(A \cap F_N) + \epsilon$ holds. (see if you can make this approach work). Another more elegant approach is done as above, using countable subadditivity to get $\leq$, then introduce a $\sup$ to get to finite $N$.

Try to forget this proof, and come up with your own. It might be fun.

The $\sigma$-algebra property.

Given a countable collection of measurable set $\Omega_j$, one need to prove that $\cup \Omega_j$ and $\cap \Omega_j$ are measurable.

We only need to prove the case of $\Omega = \cup_j \Omega_j$, since the $\cap$ operation can be obtained by taking complement and $\cup$. The hint is to define $$ \Omega_N = \cup_{j=1}^N \Omega_j$$ and $E_N = \Omega_N \RM \Omega_{N-1}$, then $\{E_j\}$ are measurable, mutually disjoint, and $\cup_j E_j = \cup_j \Omega_j = \Omega$.

Every open set can be written as a finite or countable union of open boxes.

I will leave this as discussion problem.

• A subset $U$ is open, if for every point $x \in U$, there exists an open ball $B(x,r) \In U$.
• claim: a subset $U$ is open, iff $\forall x\in U$, there exists an open box $B$, such that $x \in B \In U$.
• claim: a subset $U$ is open, iff $\forall x\in U$, there exists an open box $B$ with rational boundary coordinates, such that $x \in B \In U$.
• There are countably many open boxes with rational boundary coordinates.

All open sets are measurable.

Since open boxes are measurable, and countable union of measurable sets are measurable.

An alternative definition for measurable set is the following:

A subset $E$ is measurable, if for any $\epsilon>0$, there exists an open set $U\supset E$, such that $m^*(U \RM E) < \epsilon$.

Can you show that this definition is equivalent to the Caratheodory criterion (the one we had been using)?