Explaining maths

Although I have taught Measure and Integration many times over the years, typically I don't have time to prove all of the details of the Carathéodory theory of outer measures and their associated $\sigma$-algebras and measures. I've been thinking about how we prove some of the fiddly bits at the end, and trying to decide how to make some of the final bootstrapping from finite unions to infinite unions a bit more natural. One possibility I have thought of is to change the order in which we prove some of the results, essentially in order to avoid proving some facts twice. However, this might be at the cost of having to prove some results again for general measures that we first prove for measures coming from outer measures. On the other hand, every measure on a $\sigma$-algebra is actually the restriction to that $\sigma$-algebra of  a measure that comes from some outer measure! The same is also true for the more general "measures" on rings of sets and on semi-rings of...

At the very end of last week I was teaching the first-year mathematics students about the Extreme Value Theorem. Here is a reminder of what the theorem says. Theorem (Extreme Value Theorem) Let $a,b$ be real numbers with $a<b$, and let $f:[a,b]\to\R$ be continuous. Then there exist points $c$ and $d$ in $[a,b]$ such that, for all $x \in [a,b]$, we have $f(c)\leq f(x) \leq f(d)$. Here we have \[f(c)=\min\{f(x):x\in[a,b]\}=\inf\{f(x):x\in[a,b]\}\] and  \[f(d)=\max\{f(x):x\in[a,b]\}=\sup\{f(x):x\in[a,b]\}\,.\] Unfortunately I only had a few minutes left for this, ran out of time, and didn't finish the proof. As I am on a very tight schedule this year, I had to refer the students to the the typed notes for the rest of the proof. On the other hand, I did finish the first part of the proof, showing that a continuous real-valued function $f$ on a closed and bounded interval $[a,b]$ is always bounded above, i.e., the set $f([a,b])$ is bounded above. In this post I want to see how you ca...

 In this post we carry on from where we got to in Part I. The aim in this part is to try to tie together what we know about infimum, supremum, epsilons and sequences. This will give us a lot of flexibility in how we approach problems later: we can choose which criterion will be easiest to work with for any given situation. Although working with sequences instead of epsilon and delta can lead to some very clean proofs, you shouldn't forget about epsilon and delta! That approach is very important later throughout mathematics, and it is best to be fluent in all possible methods so that you have the most suitable method available when you need it. Let's start by having another look at supremum . The results for infimum are almost identical: you just need to flip everything e.g. by reflecting across $0$ in the number line using the map $x \mapsto -x$, which reverses all of the relevant inequalities. Let $A\subseteq \R$ be a non-empty set which is bounded above in $\R$, and set $S = ...

Let $A$ be a subset of $\R$. Then a real number $y$ is an upper bound for $A$ if $A \subseteq (-\infty,y]$, i.e., for all $a \in A$ we have $a\leq y$. We define lower bound in a similar way. Not every set has an upper bound and/or a lower bound (in $\R$). For example $\R$, $\Q$ and $\Z$ have no upper bounds and they also have no lower bounds (in $\R$). Other sets such as $\N$, $(-\infty,2)$, $[-3,5)$ have one or the other or both.  The empty set $\emptyset$ is a pathological example: every real number is both an upper bound and a lower bound for $\emptyset$, so in particular many upper bounds for $\emptyset$ are less than some of the lower bounds for $\emptyset$. Let's face it, $\emptyset$ always causes trouble! The connection between English and mathematics is not always straightforward and can lead to misunderstandings! We say that the subset $A$ of $\R$ is bounded above if there is at least one upper bound for $A$ in $\R$. Of course as soon as $A$ has an upper bound $y\in\R$,...

It is a well-known and very useful theorem of real analysis that injective continuous real-valued functions on an interval must be monotone. This is usually proved by contradiction using the Intermediate Value Theorem (IVT). However the direct negation of a function being monotone is actually not very friendly. Let $I$ be a (nondegenerate) interval in $\R$. A function $f:I\to\R$ is monotone increasing if (and only if), for all $x_1,x_2 \in I$ with $x_1\leq x_2$, we have $f(x_1) \leq f(x_2)$. Negating this we see that $f$ is not monotone increasing if there exist $x_1,x_2 \in I$ with  $x_1\leq x_2$ such that $f(x_1) > f(x_2)$.  [Of course in this case we must actually have $x_1<x_2$.] Similarly, $f$ is not monotone decreasing if and only if there exist $x_3,x_4 \in I$ with  $x_3\leq x_4$ such that $f(x_3) < f(x_4)$.  [Again, in this case we must have $x_3<x_4$.] Now $f$ is monotone (on $I$) if and only if $f$ is monotone increasing or $f$...

The Ratio Test is a powerful test for both sequences and series of non-zero real numbers. It comes in various forms, but here is one commonly used version. Theorem (Ratio Test) Let $(x_n)_{n\in\N}$ be a sequence of non-zero real numbers. Suppose that \[\frac{|x_{n+1}|}{|x_n|}\to L \text{ as } n\to\infty\,,\] where $L$ is either a non-negative real number or $+\infty$. (a) If $L>1$, then $|x_n| \to +\infty$ as $n \to + \infty$, the sequence $(x_n)$ is divergent, and the series $\displaystyle \sum_{n=1}^\infty x_n$ diverges. (b) If $L \in [0,1)$, then the series $\displaystyle \sum_{n=1}^\infty x_n$ is absolutely convergent (and hence convergent), and the sequence $(x_n)$ converges to $0$. (c) If $L=1$, then the Ratio Test is inconclusive , and you need to use a different test . (The Ratio Test tells you nothing in this case about the convergence or otherwise of either the sequence or the series.) Notes Here (b) does not tell you anything about the value of the sum of the series ...

In this post we sketch a second proof of the Hahn-Banach Extension Theorem for continuous linear functionals on real normed spaces, this time using transfinite induction. We omit some of the details, which the reader can fill in. In particular, some of the details are similar to those involved in proving the claim made in Part V (when we obtained an upper bound for a chain). For a non-zero ordinal $\gamma$, recall that, with ordinal interval notation, we have $\gamma=[0,\gamma)$. That is, $\gamma$ is equal to the set of all ordinals which are strictly less than $\gamma$. However, we use ordinal interval notation throughout. If you prefer to use $\gamma$ instead of $[0,\gamma)$ that is fine, of course! Looking at the Wikipedia page on the Hahn-Banach Theorem , apparently both Hahn and Banach (independently) used transfinite induction in their proofs, rather than Zorn's Lemma. Theorem (Hahn-Banach Extension Theorem for linear functionals on normed spaces over $\R$)  Let $(E,\jnorm)$ ...