$\gdef\vcal{\mathcal V}$

Last time we were in the middle of proving Vitali Covering Lemma, which informally says: “given a Vitali cover of a bounded measurable set $A$ using closed balls and given any $\epsilon>0$, we can find an almost cover of $A$ by countably many disjoint closed balls, where we waste no more than $\epsilon$ in the cover, and we miss only a null set in the cover”.

The construction goes by 'greedy algorithm' (which is always pick the biggest among all possible choices), however, since we have infinitely many balls, we cannot pick the biggest, so we pick some ball whose radius is 'good enough'. More precisely, we first pick an open set $W \supset A$ with $m(W) < m(A) + \epsilon$. We set $\vcal_0 = \vcal$, $W_1=W$. Then we run the iteration. starting at $n=1$

• $\vcal_n = \{ V \in \vcal_{n-1} | V \In W_n \}$, then $\vcal_n$ is still a Vitali cover of $A \cap W_n$ (why?)
• $d_n = \sup{diam V | V \in \vcal_n\} $
• choose $V_n$ so that $diam (V_n) > d_n/2$.
• set $W_{n+1} = W_n \RM V_n$.
• increase $n$ by 1 and repeat

After we run the algorithm and obtain a collection of disjoint balls $V_n$, we need to show that these balls cover $A$ up to a null set. Last time, we proved that, for any positive integer $N$, we have $\cup_{k \geq N} 5 V_k \supset A \RM (V_1 \cup \cdots \cup V_{N-1})$ (recall the proof)

Why is this useful? It allows us to say, for any $\delta > 0$, there exists an $N$, so that $m(A \RM (V_1 \cup \cdots \cup V_{N-1}) ) < \cup_{k \geq N} 5 V_k) < 5^n \sum_{k=N}^\infty m(V_k) < \delta$. (This last inequality is always achievable by choosing$N$large enough, since$\sum_{k=1}^\infty m(V_k) < m(W) < \infty$. )

We decompose $\R^n$ into a grid of size $1$ cubes, throw aways the boundaries (which is a measure zero set). We enumerate these cubes as $\{C_n\}$, then for any $\epsilon>0$, we find Vitali cover for $A \cap C_i$ with excess $\epsilon / 2^i$. Then we union together the solution to the sub-problemes (countable union of countable collection is still a countable collection).

General shape of the ball. Actually, we can work with non-Euclidean norm. (can the shape be more general?)

Suppose $E \in \R^n$ is measurable, for any $p \in \R^n$, we define density of $E$ at $p$ to be $\delta(p,E) = \lim_{Q \downarrow x} m(E \cap Q)$ remark:

• here we don't have a sequence, so what does convergence mean? (limit indexed by a poset)
• we could define lower density $\underline\delta(p,E)$ using $\liminf$; similarly upper density..

Example, if $E = (0,1) \In \R$, does density exists at the boundary of $E$?

If $\delta(p,E)=1$, we say $p$ is a density point of $E$.

Lebesgue density theorem. Almost all points of $E$ are density point.

Proof: define for any $0 \leq a < a$, let $E_a = \{p \in E \mid \underline\delta(p,E) < a \}$. Claim $m^*(E_a) =0$. Given the claim, then $E_{<1} := \cup_{0 \leq a<1} E_a = \cup_{n=1}^\infty E_{1/n}$ is null set. And any nondensity point $p$ would belong to $E_{<1}$, then we are done.

To prove the claim $m^*(E_a)=0$, we take the collection of cubes $Q$, such that $[Q: E] = m(Q \cap E) / m(Q) < a$. Such cubes form a Vitali covering of $E_a$, indeed, for any $p \in E_a$ and any $r >0$, there are some cube $Q$ contained in $B_r(p)$ (containing p), with $[Q:E] < a$. Then, using Vitali covering by such cubes, and for any $\epsilon>0$, we can get disjoint almost cover of $E_a$ by such $Q$ with margin size $\epsilon$. Then we have $$ m^*(E_a) = \sum_{i=1}^\infty m^*(E_a \cap Q_i) \leq \sum_{i=1}^\infty m^*(E \cap Q_i) \leq \sum_{i=1}^\infty a m^*(Q_i) = a \sum_i m(Q_i) \leq a(m(E_a) + \epsilon) $Thus,$m^*(E_a) \leq (a/ (1-a)) \epsilon$, for any$\epsilon$, hence$m^*(E_a)=0$.

$\gdef\uint{\overline{\int}}$ $\gdef\lint{\underline{\int}}$

• Redo Fubini's Theorem (Tao 8.5.1)
• Vitali Covering Lemma.

To deal with possibly non-integrable functions, we need to define 'upper Lebesgue integral' and 'lower Lebesgue integral', which works for non-integrable functions $\overline{\int} f(x) = \inf \{ \int g(x), \text{g absolutely integrable, and $g(x)>f(x)$} \}$ similarly for lower Lebesgue integral. By monotonicity of integral, we always have upper integral greater than lower integral.

Lemma 8.3.6 says, if a function $f: \R^n \to \R$ satisfies $\uint f = \lint f$, then $f$ is absolutely integrable. To prove it, we create a sequence that approximate $f$ from above, $\overline f_n$ and a sequence that approximate $f$ from below $\underline f_n$, take their limit to get $F_+, F_-$ with $F_+ \geq F_-$. Since $\int F_+ = \uint f = \lint f = \int F_-$, we have $\int F_+ - F_- = 0$, since $F_+ -F_-\geq 0$, we have $F_+ = F_-$ a.e., since $F_+ \geq f \geq F_-$, thus $f = F_+$ a.e., thus measurable and absolutely integrable.

Let $f(x,y): \R^2 \to \R$ be an absolutely integrable function, then there exists integrable function $F(x)$ and $G(y)$, such that for a.e $x$, we have $F(x) = \int f(x,y) dy$ and for a.e $y$, $G(y) = \int f(x,y) dx$, and $\int f(x,y) dx dy = \int F(x) dx = \int G(y) dy$

Pf: We only consider the statement about $F(x)$.

1. We introduce two non-negative functions $f_-, f_+$ (with disjoint support) such that $f = f_+ - f_-$. Then, suppose we prove the theorem for $f_+$ and $f_-$, we can combine the result to get that of $f$. Thus, we only need to deal the case where $f$ is non-negative.
2. Since $f$ is the sup of functions with bounded support, $f = \sup_N f_N, f_N = f \chi_{ [-N, N] \times [-N, N]}$, if $F_N(x) = \int f_N(x,y) dy$ for $x \notin Z_N$, then then we can define $F = \sup_N F_N$, For $x \notin Z = \cup_N Z_N$ (countable union of null-set is still null), by monotone convergence theorem (for upward non-negative functions, we have $F(x) = \sup_N F_N(x) = \sup_N \int f_N(x,y) dy = \int \sup_N f_N(x,y) dy = \int f(x,y) dy$ Hence, suffice to prove the statement for functions with support in a big box $[-N, N] \times [-N, N]$.
3. By since $f$ is $\sup$ of simple functions, we may replace $f$ by simple functions, and go back to $f$ using monotone convergence theorem.
4. Replace simple function by characteristic function, by linearity of integration.
5. Here is the core, we only need to prove the case for $f = 1_E$, where $E \In [-N,N]^2$ is a measurable set. We claim that $\uint (\uint 1_E(x,y) dy) dx \leq m(E)$. The proof of the claim is by box covering, as we did last time (see Tao for detail). Given this claim, we can now finish the proof. Let $E^c = [-N,N]^2 \RM E$, then we have

$4N^2 - \lint (\lint 1_E(x,y) dy )dx = \uint (\uint 1_{E^c}(x,y) dy) dx \leq m(E^c) = 4N^2 - m(E)$ So, $\lint (\lint 1_E(x,y) dy )dx \geq m(E)$ In particular, $\lint (\uint 1_E(x,y) dy )dx \geq \lint (\lint 1_E(x,y) dy )dx \geq m(E) \geq \uint (\uint 1_E(x,y) dy )dx \geq \lint (\uint 1_E(x,y) dy )dx$ Hence $F_+(x) = \uint 1_E(x,y) dy$ is integrable. Similarly $\lint (\lint 1_E(x,y) dy )dx \geq m(E) \geq \uint (\uint 1_E(x,y) dy )dx \geq \uint (\lint 1_E(x,y) dy )dx \geq \lint (\lint 1_E(x,y) dy )dx$ thus $F_- (x) = \lint 1_E(x,y) dy$ is integrable. And, we have $\int F_+(x) dx = \int F_- (x) dx$ hence $F_+(x) = F_-(x)$ for almost all $x$. Thus, for a.e. $x$, we have $\uint f(x,y) dy = \lint f(x,y) dy$, thus $\int f(x,y) dy$ exists for a.e. x.

Suppose $A$ is measurable, and $B \In A$ any subset, with $B^c = A \RM B$. Then $m(A) = m^*(B) + m_*(B^c)$ Proof: $m^*(B) = \inf \{ m(C) \mid A \supset C \supset B, C \text{measurable} \} = \inf \{ m(A) - m(C^c) \mid A \supset C \supset B, C \text{measurable} \}$ $= m(A) - \sup \{ m(C^c) \mid A \supset C \supset B, C \text{measurable} \} = m(A) - \sup \{ m(C^c) \mid C^c \subset B^c, C^c \text{measurable} \} = m(A) - m_*(B^c)$

$\gdef\vcal{\mathcal{V}}$ A Vitali covering $\vcal$ of a set $A \In \R^n$ is such that, for any $p \in A, r > 0$, there is a covering set $V \in \vcal$, such that $\{p\} \subsetneq V \In B_r(p)$, where $B_r(p)$ is the open ball of radius $r$ around $p$.

Vitali Covering Lemma: Let $\vcal$ be a Vitali covering of a measurable bounded subset $A$ by closed balls, then for any $\epsilon>0$, there is a countable disjoint subcollection $\vcal' = \{V_1, V_2, \cdots \}$, such that $A \RM \cup_k V_k$ is a null set, and $\sum_k m(V_k) \leq m(A) + \epsilon$.

Proof: The construction is easy, like a 'greedy algorithm'. First, using the given $\epsilon$, we find an open subset $W \supset A$, with $m(W) \leq m(A) + \epsilon$. Let $\vcal_1 = \{V \in \vcal: V \In W\}$, and $d_1 = \sup \{diam V: V \in \vcal_1\}$. We pick $V_1 \in \vcal_1$ where the diameter is sufficiently large, say $diam V_1 > d_1 /2$. Then, we delete $V_1$ from $W$, let $W_2 = W \RM V_1$, and consider $\vcal_2 = \{ V \in \vcal_1, V \In W_2\}$, and define $d_2 = \sup \{diam V: V \in \vcal_2\}$, and pick $V_2$ among $\vcal_2$ so that $diam V_2 > d_2 /2$. Repeat this process, we get a collection of disjoint closed balls $\{V_i\}$. Suffice to show that $A \RM \cup V_i$ is a null set.

The crucial claim is the following, for any positive integer $N$, we have $\cup_{k=N}^\infty 5 V_k \supset A \RM (\cup_{i=1}^{N-1} V_i)$ Suppose not, and there is a point $a \in A \RM (\cup_{i=1}^{N-1} V_i)$, but not in $\cup_{k=N}^\infty 5 V_k$, then we can find a closed ball $B \in \vcal_N$, such that $a \in B$. Since $a \notin 5V_N$, we have $B \not\subset 5V_N$. This implies $B \cap V_N = \emptyset$. Draw a picture. This implies $B \in \vcal_{N+1}$. Then, repeat the above story $N$ replaced by $N+1$ and same $a,B$, we can keep going and show that $B \in \vcal_k$ for all $k \geq N$. That cannot be true, since $d_k \to 0$, but $B$ has fixed radius.

Today we covered Tao 8.3, 8.4, and 8.5. Here is the video, but I made a stupid mistake regarding Fubini theorem.

I made a mistake in today's presentation in 8.5. Namely, given a measurable function $f(x,y)$. First of all, for a fixed $x$, the function $f_x(y) = f(x,y)$ as a function of $y$, may not be measurable at all. For example, take a measurable subset $E \In \R^2$, it is possible that certain slice $E_x = E \cap \{x\} \times \R$, when viewed as a subset of $\R$, is non-measurable (it is measurable as a subset of $\R^2$, a null-set), then consider $f$ as indicator function $1_E(x,y)$. Hence, the proper way to state the Fubini theorem, is that, there exists a measurable function $F(x)$, such that there exists a null-set $Z$, and for $x \notin Z$, we have $f_x(y)$ is measurable, and $F(x) = \int f_x(y) dy.$ and $\int F(x) dx = \int f(x,y) dx dy$

I will revisit this theorem on Thursday.

We did Tao 8.2.

Main result is monotone convergence theorem: given a monotone increasing sequence of non-negative measurable functions $f_n$, we have $\int \lim f_n = \lim \int f_n$ or equivalently $\int \sup f_n = \sup \int f_n$ The $\geq$ direction is easy, the $\leq$ direction is hard, which requires 3 steps lowering of the LHS $\int \sup f_n$:

• We first replace $\sup f_n$ by simple functions $s$, with $\sup f_n \geq s$, for some simple function $s$ sub-ordinate to $\sup f_n$.
• We then lower $s$ a bit, $s \geq (1-\epsilon)s$.
• We then cut-off the integration domain a bit, by introducing a cut-off function $1_{E_n}(x)$, where $E_n= \{x: (1-\epsilon)s(x) \leq f_n(x) \}$, we get $(1-\epsilon)s \geq (1-\epsilon) s 1_{E_n}$.

After the three lowering, we get $(1-\epsilon) s 1_{E_n} \leq f_n$, hence $\int (1-\epsilon) s 1_{E_n} \leq \int f_n \leq \sup \int f_n$ Then, we reverse the above lowering process, by taking limit, or sup over all possible choices

• First, we let $n \to \infty$. By proving directly a 'baby version' of monotone convergence theorem for simple functions, we have that $\sup \int s 1_{E_n} = \int s \sup 1_{E_n} = \int s$. This gives us

$\int (1-\epsilon) s \leq \sup \int f_n$

• Then, we take limit $\epsilon \to 0$, to get $\int s \leq \sup \int f_n$
• Finally, we sup over all simple functions $s$ subordinate to $\sup f_n$, to get $\int \sup f_n \leq \sup \int f_n$

Then, we did some applications. For example, summation and integration can commute now (for non-negative measurable functions).

We will cover Tao's 7.5 and 8.1 today. Here we will use Tao's definition of measurable set, and Lebesgue integration, which a priori is not the same as Pugh's.

Let $\Omega \In \R^m$ be measurable, and $f: \Omega \to \R^m$ be a function. If for all open sets $V \In \R^m$, we have $f^{-1}(V)$ being measurable, then $f$ is called a measurable function.

If $f: \Omega \to \R^m$ is continuous, then $f$ is measurable. Indeed, if $f^{-1}(V)$ is open in $\Omega$, then $f^{-1}(V)$ is an intersection of open subset $U \In \R^m$ and $\Omega$ (recall the definition of topology on $\Omega$), an intersection of two measurable sets.

Instead of checking on all open sets $V \In \R^m$, we can just check for all open boxes in $\R^m$. Since any open can be written as a countable union of open boxes.

A measurable function $f$, post compose with a continuous function $g$ is still measurable. Since $(g \circ f)^{-1}(open) = f^{-1} (g^{-1}(open)) = f^{-1}(open) = measurable$

Lemma: $f: \Omega \to \R$ is measurable if and only if for all $a \in \R$, $f^{-1}( (a, \infty))$ is measurable.
Proof: every open set in $\R$ is a countable union of open interval, hence suffice to show that all open intervals $(a,b)$ has pre-image being measurable. We can easily show that $f^{-1}((a, b])$ is measurable for all $a<b$, and we can use countable operations to approximate open interval by half-open-half-closed ones, $(a,b) = \cup_{n} (a, b-1/n]$.

Simple functions are measurable functions $f: \Omega \to \R$, which takes value in a finite subset of $\R$.

Simple functions forms a vector space (i.e., closed under addition and scalar multiplication), and can be written as a finite linear combination of characteristic functions $\chi_E$.

The important thing is that, any non-negative measurable function $f$ admits a sequence of simple functions $f_n$, non-negative, and $f_n \leq f_{n+1}$, such that $f_n \to f$ pointwise. The construction requires both 'trunction' and refinement.

We then define integration for simple functions. Integration is a linear map from the vector space of simple function to $\R$.

Finally, in 8.2, we will define integration for non-negative measurable functions.

$\int f = \sup \{ \int s \mid 0 \leq s \leq f, \text{$s$ is a simple function \} $

For $f,g : \Omega \to [0, \infty]$, how to prove $\int f+g = \int f + \int g$?