Notes
Lecture 3
Lemma 7.4.2
I will prove the easy case (1-dim), you will do the general case in HW.
Let $E = (0,+\infty)$ be the open half space in $\R$. For any subset $A \In \R$, we need to prove that $$ m^*(A) = m^*(A \cap E^c) + m^*(A \cap E). $$ Let $A_+ = A \cap E$ and $A_- = A \cap E^c$, note that if $0 \in A$, then $0 \in A_-$. First, we note that $$ m^*(A) \leq m^*(A_-) + m^*(A_+)$$ by finite sub-additivity of outer-measure. (This is because any open cover of $A_-$ and open cover of $A_+$, union together form an open cover of $A$.) Next, we need to show that, for any $\epsilon>0$, we have $$ m^*(A) + 2\epsilon \geq m^*(A_-) + m^*(A_+).$$ The plan is the following. Given an open covering of $A$ by intervals $\{B_j\}_{j=1}^\infty$, such that $m^*(A) + \epsilon > \sum_j |B_j|$. We define $$ B_j^+ = B_j \cap (0,\infty) , \quad B_j^- = B_j \cap (-\infty, \epsilon / 2^j) $$ then $\{B_j^+\}$ forms a cover of open interval of $A_+$, similarly $\{B_j^+\}$ forms a cover of open interval of $A_-$. $|B_j^+| + |B_j^-| \leq |B_j| + \epsilon / 2^j$. Thus, we have $$ m^*(A) + 2\epsilon \geq \sum_j |B_j| + \epsilon \geq \sum_j |B_j^+| + \sum_j |B_j^-| \geq m^*(A_+) + m^*(A_-). $$
That finishes the proof for $n=1$ case. How to generalize to higher dimension? In the above discussion, we used the trick of $1 = \sum_j 1/2^j$, which is same trick as we prove outer-measure has countable sub-additivity. See the hint in Tao's Exercise for a two-step prove, that utilizes many results that we have proven.
Lemma 7.4.4
We only prove some part here.
- © If $E_1, E_2$ are measurable, then $E_1 \cup E_2$ and $E_1 \cap E_2$ are measurable.
- (e) every open box, and every closed box, is measurable
- (f) if a set has outer-measure zero, then it is measurable.
Proof:
- © Try the case first with $A = \{ x^2 + y^2 < 1 \} \In \R^2$ and $E_1 = \{x>0\}$ and $E_2 = \{y>0\}$, then we can partition $$ A = A_{++} \sqcup A_{+-} \sqcup A_{-+} \sqcup A_{–} $$ (cut the pie to 4 pieces), then we have finite additivity of outer-measure in this case. Such conclusion holds in general, assuming $E_1, E_2$ are measurable.
- (e) express open box as intersections of translated half-spaces.
- (f) Assume $E$ has outer-measure 0, then $0 \leq m^*(A \cap E) \leq m^*(E) = 0$, we can prove $m^*(A) \geq m^*(A \RM E) + m^*(A \cap E) = m^*(A \RM E)$ again by monotonicity.
Lemma 7.4.5
finite additivity: if $\{E_j\}_{j=1}^N$ is a finite collection of disjoint measurable sets, then for any subset $A$, we have $$ m^*(A \cap (\cup_j E_j)) = \sum_j m^*(A\cap E_j) $$
Proof: again, we try the case of $N=2$ first, to get intuition. Let $B = A \cap (E_1 \cup E_2)$, and $B_1 = A \cap E_1 = B \cap E_1$, and $B_2 = A \cap E_2 = B\cap E_2$, so, we are trying to prove that $$ m^*(B) = m^*(B_1) + m^*(B_2) $$ since $E_1$ and $E_2$ are disjoint, we notice that $B_2 = B \RM E_1$, thus the above holds by measurability of $E_1$ applied to the test set $B$
Discussion Time
Let's fill in the details of the above sketches.
Lecture 2
Last time we had the definition of outer measure, and we basically followed Tao-II's presentation. This time, we will go through Lemma 7.2.5 (relatively easy) and Lemma 7.2.6 (about outer measure of a box, a bit hard). Pugh gives a different proof for the outer measure of a box being what it supposed to be, namely the naive volume, and he uses Lebesgue number. I am going to follow Tao's approach, although it is longer.
Then, we plan to talk about the construction of 'non-measurable set', in Tao-II, 7.3. And then, give the definition of measurable set, that follows the Caratheodory condition. There is an alternative and equivalent definition, see Tao-M (Tao measure theory grad textbook), which says $E \In \R^n$ is measurable, if for any $\epsilon>0$, there exists an open set $U \supset E$, such that $m^*(U\RM E) < \epsilon$, namely, measurable set are those than can be approximately from the outside by an open set.
We will use the discussion time, hopefully 30 minutes, to tackle Lemma 7.4.2, Lemma 7.4.4.
Lecture 1
- Motivation for Lebesgue measure (read Tao 7.1)
- What is the definition of outer-measure?
- Hmm, the outer-measure of a closed box? Why is it so complicated?
- Interlude, the Banach-Tarski sphere.
Discussion Time:
- How to prove that $\{1,2,3\}$ has zero outer-measure?
- How to prove that $\Z$ has zero outer-measure? And $\Q$?
- Can you summarize some rules that allow you get to the above results quicker?
- Can you show that $\R$ has zero outer-measure in $\R^2$? Can you prove that in general, a lower dimensional 'manifold' in $\R^n$ has zero outer measure? Like the unit circle in $\R^2$?
- (hard) How does outer-measure behave under product? Does $m^*(A \times B) = m^*(A) \times m^*(B)$?
Sketch
Welcome to this class. So, I assumed you all had taken math 104 or the equivalent of it, which covers sequence and limits, metric space topology (open sets, distance functions, compact sets etc), and also some Riemann integrals. Why do we want to take the second course in analysis?
Here is what this course is mainly about: Lebesgue measure theory and integration, then some multivariable calculus (will be useful for differentiable manifold) and Fourier analysis (what functions can be approximated by sums of sine and cosine?).
Why do we need to use Lebesgue integral, rather than Riemann integral? Why isn't knowing the length of an interval $[a,b]$ to be $b-a$ enough? Before we say Riemann integral is not good enough, let's first recall what it is good for. Piecewise continuous functions are Riemman integrable, in particular piecewise constant function are Riemann integrable. But it is very limited. In particular, pointwise limit of Riemann integrable function may not be Riemann integrable (though uniform limit preserves it).
The problem is that, there are sometimes subsets in $\R$, which are not interval, but infinite unions of it. Our dream is that, given any subset $A$ of $\R$, we can assign to it a number, the 'measure' of $A$, that 'reflect' its size. Now, that is too vague. What do we mean when we say reflect its size? It better have some nice properties, like
- monotonicity: if $A \In B$, then $m(A) \leq m(B)$.
- additivity: if $A = B \sqcup C$, $A$ is the disjoint union of $B$ and $C$, then $m(A) = m(B) + m(C)$.
- translation invariance: for any $x \in \R$, such that $m(x+E) = m(E)$.
It turns out, it is impossible to find such a measure function on all subset, but it is only possible to define it for a sufficiently nice subset, which we will call, 'measurable subset'. There are certain desirable properties we want to have on measurable sets (they form a sigma-algebra, and contains all open subsets).
Let's pause and recall the definition of Boolean and Sigma-algebra. In computer science, you may have heard about the opeartion (not, and, or), defined on a set, here we have exactly the same notion, but in different notation, 'not'='complement', 'and'='intersection', 'or'='union'.
Ex: show that, if $S = \{a,b,c\}$, then the power set $2^S$ (the set of all subsets of $S$) consist of 8 elements, corresponding to $2^3=8$. Work out how the correspondence go.
Then, what is the sigma algebra? Sigma algebra is a stronger requirement than the Boolean algebra, since it allows for 'countably' many operations taken at a time.
:?:Why the word 'countable' is important? Can you replace it by the more general word 'infinite'?
Our plan in the following, is to first define the notion of an outer-measure $m^*(A)$, that works for all subset $A \In \R^n$. It has many nice properties, which we will spend this lecture and next explore. But, it is not a measure, because it fails the additivity condition. Then, we introduce the notion of Lebesgue measurability.
First, we define the measure of an open 'box' in $\R^n$. Then, for any given subset $A$, we cover it by countably many open boxes, and get an approximation of the outer-measure, and we optimize over the covering.
Lemma 7.2.5 as discussion problem. The trick, is to given oneself an epsilon of room, and cut $\epsilon$ into countably many small pieces $\epsilon = \epsilon \sum_{n=1}^\infty 1/2^n$.
Now, here is a tricky pit-fall. Given the definition of an outer-measure, how to compute the outer-measure of an open box? Is it possible that we use some trickery, we can cover a box using smaller boxes with a smaller total volume?
Let's see a few examples.
- Outer measure of $\Q$ in $\R$?
- Outer measure of $\R$ in $\R^2$?
