Table of Contents
Selena Li
Hi! I'm Selena, an Applied Math & Statistics double major.
Lecture Notes
Homework
Final Essay
Here are some notes I wrote on probability measure spaces for my final essay.
Other Notes/Thoughts
Uncountable Coverings
A definition of an infinite sum over an uncountable set was brought up in the discord (thanks Max): $$\sum_{i \in I} r_i = \sup \{ \sum_{i\in J} r_i | J\subseteq I, J \textrm{ finite} \}$$
The definition of outer measure restricts to countable coverings. However, it appears that this restriction is not strictly necessary. Suppose we had an uncountable box covering of a set $A$. Since boxes have non-zero volume, we have uncountably many non-zero values in our sum, implying a sum of infinity. Because we can always find a countable box covering of $A$ (in particular, we can cover all of $\mathbb{R}^n$ using the unit boxes with vertices at integer coordinates) with volume $\leq \infty$, the set of countable box coverings determines the infimum and relaxing the constraint to allow for uncountable coverings effectively changes nothing.
$\mathbb{R}$ has outer measure zero?
Since $\mathbb{Q}$ is countable, using the $\frac{\epsilon}{2}$ trick, for any $\epsilon > 0$, we can construct a countable union of open intervals of total length less than $\epsilon$ with an interval centered at $q$ for each $q\in \mathbb{Q}$.
Intuitively, it seems that such a covering would also cover all the real numbers, thus implying that $m^*(\mathbb{R}) =0$, which is false. Suppose $q \in \mathbb{Q}$ has corresponding interval $(q-\delta, q+\delta)$. Since we know there is rational between every pair of reals, there is some $q' \in \mathbb{Q} \cap (q, q+\delta)$. Thus, we have successfully covered all the real numbers between $q$ and $q'$. Then, you could repeat this with $q'$, covering all the real numbers between $q'$ and some larger rational, etc. until you covered the whole real line.
Clearly, this doesn't quite work because you could end up with sequence of interval lengths whose sum converges to a finite number, and you'd never make it past a certain point on the real line. But it still feels like you're going to cover all the gaps between the rationals, and the real numbers are contained in these gaps.
One explanation I came up with for why this intuition is wrong is because this notion of having “gaps” seems to assume that there's some order for the rational numbers. Although countability implies that we can number the rationals, we still can't put them in order along the real line–if you had some order $q_1, q_2, …$, then $\frac{q_1+q_2}{2}$ would also be rational, but it belongs between $q_1$ and $q_2$.
Furthermore, say you had some $r\in \mathbb{R}$. How would you check that $r$ is covered by by some interval? Well, you could pick some rational $q$ that is close to $r$. If the interval around $q$ is doesn't contain $r$, we can move over to some interval around a rational that is closer to $r$ than $q$, and repeat. But as before, the sequence of rationals could converge before you actually make it to $r$.
Just some random thoughts I've been having. The whole thing still feels rather counterintuitive, so if anyone has any input or a way of visualizing what a covering of $\mathbb{Q}$ looks like, I would love to hear it.
Lesbesgue Integral
We define the $\textbf{undergraph}$ of a function $f: \mathbb{R} \rightarrow [0, \infty)$ to be the set $$\mathcal{U}f = \{(x, y) \in \mathbb{R}: 0 \leq y < f(x)\}$$
Then, $f$ is $\textbf{Lebesgue measurable}$ if $\mathcal{U}f$ is measurable, and we define its $\textbf{Lebesgue integral}$ to be $$\int f = m (\mathcal{U}f)$$
Finally, if $\int f$ is finite, $f$ is $\textbf{Lebesgue integrable}$. Note that we can also restrict to a smaller domain $X \subseteq \mathbb{R}$ by looking at a subset of the undergraph $$\mathcal{U}_Xf = \{(x, y) \in X: 0 \leq y < f(x)\}$$
One difference between the Riemann and Lebesgue integrals is that the Riemann integral is only defined for bounded functions on bounded domains. Since the Lebesgue integral is not subject to these restrictions, it opens up the possibility of integrating over a wider range of functions. A classic example is the function $\chi_{\mathbb{Q}}$, which takes on the value $1$ at all the rationals and $0$ at all the irrationals. Thus, we see that the Lebesgue integral is much more general than the Riemann integral.
Lebesgue Measure Theory Summary
Tao's approach to Lebesgue measure theory starts by defining a $\textbf{simple function}$, which takes on finitely many possible values. We first define how to integrate a simple function, and then use that to define how to integrate measurable functions.
Simply put, the integral of a simple function is defined to be the product of each possible value with the measure of the subset of the region of integration that corresponds to the possible value. Then, to integrate non-negative measurable functions, we define the integral of a measurable function f to be the supremum of the integral of non-negative simple functions that are less than or equal to f.
Some of the key results for Lebesgue integration of non-negative functions include:
$\textbf{Lebesgue Monotone Convergence Theorem}$: For a sequence $(f_n)$ of functions from $\Omega \rightarrow \mathbb{R}$, where $\Omega \subseteq \mathbb{R}^n$ is measurable and $\forall x\in \Omega$, $$0 \leq f_1(x) \leq f_2(x) \leq …$$ we have that $$0 \leq \int_{\Omega} f_1(x) \leq \int_{\Omega} f_2(x) \leq …$$ and $$\int_{\Omega} \sup{f_n} = \sup{\int_{\Omega} f_n}$$
The proof of the inequality in the $\geq$ direction is simple. For the opposite direction, we first use the definition of Lebesgue integration to look at simple functions less than or equal to $\sup{f_n}$. Then, we use the trick where we show the inequality holds after cutting by a factor of an arbitrarily small epsilon. Finally, after cutting our simple function $s$ to $(1-\epsilon)s$, we integrate over $E_n \subseteq \Omega$, where $E_n$ is the set of values where $f_n$ is greater than $(1-\epsilon)s$. Observing that $\int_{E_n}(1-\epsilon)s \leq \int_{\Omega} f_n$ and taking the supremum completes the proof. In summary, the key idea used in this proof is that to show an inequality, we can cut by arbitrarily small amounts in multiple places, and taking the as $\epsilon$ tends to zero shows the result.
$\textbf{Interchange of Integration and Addition}$: $\int_{\Omega} (f+g) = \int_{\Omega} f + \int_{\Omega} g$
Tao's proof of this theorem is easier to follow (at least in my opinion) than Pugh's. The main idea here is to construct increasing sequences of simple functions approaching $f$ and $g$, and apply results for simple functions and take the supremum.
$\textbf{Fatou's Lemma}$: For a sequence $(f_n)$ of functions from $\Omega \rightarrow \mathbb{[0, \infty]}$, where $\Omega \subseteq \mathbb{R}^n$ is measurable, $$\int_{\Omega} \lim_{n \rightarrow \infty} \inf{f_n} \leq \lim_{n\rightarrow \infty} \inf{\int_{\Omega} f_n}$$
The key idea in the proof is rewriting the liminf on the LHS as a sup, and apply the Monotone Convergence Theorem.
Finally, for integration of measurable functions that are not necessarily non-negative, Tao introduces the idea of an $\textbf{absolutely integrable}$ function, which is a function whose absolute value is Lebesgue integrable. Also, we introduce the positive and negative part of a function $f$, defined by $$f^+ = \max{(f, 0)}, f^- = -\min{(f, 0)}$$
Then, we define $$\int f = \int f^+ - \int f^-$$
Many of the properties of the integral of non-negative functions can be generalized to the general Lebesgue integral.
A key theorem involving absolutely integrable functions is:
$\textbf{Lebesgue Dominated Convergence Theorem}$: Let there be a sequence $(f_n)$ of functions from $\Omega \rightarrow \mathbb{R}^*$ converging pointwise, where $\Omega \subseteq \mathbb{R}^n$ is measurable. If $\exists F: \Omega \rightarrow [0, \infty]$ s.t. $|f_n| \leq F $ for each $n$, $$\int_{\Omega} \lim_{n\rightarrow\infty} f_n = \lim_{n\rightarrow\infty}\int_{\Omega}f_n$$
To prove this theorem, we use Fatou's Lemma on the sequences $F + f_n$ and $F - f_n$, both of which are non-negative from the condition. This allows us to lower bound $\int_{\Omega} f$ by the limsup of $\int_{\Omega} f_n$ and upper bound it by the liminf, which shows the desired result.
Littlewood's Three Principles
Littlewood's Three Principles summarize three main ideas in real analysis. The first principle says that every measurable set can be written as the union of finitely many intervals, up to a set of measure $\epsilon$. The second principle says that every measurable function is continuous on it's domain minus a set of measure $\epsilon$. The third principle says that a pointwise convergent sequence of measurable functions converges uniformly up to an $\epsilon$-set.
These three principles share a common theme that removing an arbitrarily small set can lead to nice properties, or that measurable functions/sets “nearly” have these properties.
