Typical elements 2

Since we’ve already discussed the use of simple element in the proof of the ergodic theorem, we will now address the existence of such elements.

Recall that we are working with a measure preserving system ([0,1]^{\mathbb{N}}, \mathcal{C}, T, \nu) as defined previously, where T denotes the shift operator. We are looking for an element \alpha of [0,1]^{\mathbb{N}} such that

\displaystyle \lim_{n\to \infty} \sum_{i=0}^{n-1} f(T^i \alpha ) = \int_{[0,1]^{\mathbb{N}}} f(y)d\nu .

for any f \in C([0,1]^{\mathbb{N}}). First, take a sequence \{ \alpha_n \}_{n \in \mathbb{N}} in [0,1]^{\mathbb{N}} of elements  that are periodic; in other words, each element is of the form

\displaystyle \alpha_n = (\alpha_{1}^{n} , \alpha_{2}^{n} , \dots , \alpha_{c_n }^{n} , \alpha_{1}^{n} , \alpha_{2}^{n} , \dots, \alpha_{c_n }^{n} , \dots )

for some integer c_n. We set

\displaystyle \mu_{\alpha_n } = \frac{1}{c_n } (\delta_{\alpha_n } + \delta_{T  \alpha_n } + \cdots + \delta_{T^{c_{n-1} }\alpha_n } ),

where  \delta denotes the Dirac measure, as usual. We require that the \mu_{\alpha_n} converge weakly to \nu as n\to \infty.

Here we depart from the sequence of the proof in Kamae in order to first show how to construct such a sequence, before showing that we can use it to obtain a typical element. To show the existence of the sequence, it suffices to show that for any neigbourhood U of \nu in the weak topology of Borel measures on [0,1]^{\mathbb{N}}, we can find an element \beta such that \mu_{\beta}\in U. The nature of the topology means that we can approximate the measure \nu to an arbitrary degree by considering a finite amount of information. In other words, there are integers m, n and some positive \delta >0 such that if | \mu (B) - \nu (B) |<\delta   for any subset B of the form

\displaystyle \left[ \frac{j_0}{m} , \frac{j_0 +1}{m} \right] \times \cdots \times \left[ \frac{j_{n-1}}{m} , \frac{j_{n-1}+1}{m} \right] \times [0,1] \times [0,1] \times \cdots

where j_0 , j_1 ,\dots ,j_{n-1} are integers ranging from 0 to m-1, then \mu \in U.

Now set

\displaystyle \Sigma = \left\{ \frac{1}{2m} , \frac{3}{2m} , \dots , \frac{2m-1}{2m} \right\} .

We can define a measure \lambda on \Sigma^n by

\displaystyle \lambda \left( \left( \frac{j_{0} + \frac{1}{2}}{m} , \dots , \frac{j_{n-1} +\frac{1}{2}}{m} \right) \right)  = \nu \left( \left[ \frac{j_{0} }{m} , \frac{j_{0} +1}{m} \right]  \times \dots \times \left[ \frac{j_{n-1} }{m} , \frac{j_{n-1} +1}{m} \right] \times [0,1] \times \dots \right)

for the j_i in the same range as before. We can see \lambda as an encoding of the measure \nu on cylinder sets of the form we are considering, and will clearly be used to construct a measure in the neighbourhood of \nu, once we have tweaked it a little.

We are given that \nu is T-invariant. From this, we can conclude that

\displaystyle \sum_{\xi_0 \in \Sigma} \lambda ((\xi_0 , \chi_1 ,\dots , \xi_{n-1} )) =\sum_{\xi_0 \in \Sigma} \lambda ((\xi_1 , \dots , \xi_{n-1} , \xi_{0} )). \quad  (*)

for any \xi_1 , \dots \xi_{n-1} \in \Sigma. Why? We have that

\displaystyle \bigcup_{\xi_0  \in \Sigma}  \left[ \frac{j_0}{m} , \frac{j_0  +1}{m} \right] \times \cdots \times \left[ \frac{j_{n-1} }{m} , \frac{j_{n-1} +1}{m} \right] \times [0,1]  \times [0,1]  \times \cdots \\ = T^{-1} \left(  \left[ \frac{j_1 }{m} , \frac{j_1 +1}{m} \right] \times \cdots \times \left[ \frac{j_{n-1} }{m} , \frac{j_{n-1} +1}{m} \right] \times [0,1] \times [0,1] \times \cdots   \right) ,

and the latter term is equal to

\displaystyle \bigcup_{\xi_0  \in \Sigma} \left( \left[ \frac{j_1 }{m} , \frac{j_1 +1}{m} \right] \times \cdots \times \left[ \frac{j_{n-1} }{m} , \frac{j_{n-1} +1}{m} \right] \times \left[ \frac{j_0 }{m} , \frac{j_0 +1}{m} \right] \times [0,1]  \times \cdots \right) .

From \lambda, we want to obtain a measure \nu on \Sigma^n which satisfies:

(1) For any \xi \in \Sigma^n, N\eta (\xi) is a positive integer such that |\lambda (\xi) - \eta (\xi) |<\delta, where \delta has been specified earlier.

(2) The relation (*) holds for \eta in the place of \lambda.

These do not seem like very great demands on \eta, but we should still confirm that it can satisfy them. The first condition above requires that all the values \eta (\xi) are rational. This is not much of a challenge, because of the denseness of the rationals. We must just make sure that the adjustment from \lambda to \eta does not violate condition (1). We therefore need to make sure, also, that the rationals we pick not only satisfy (*), but also

\displaystyle \sum_{\xi_0 , \xi_1 , \dots , \xi_{n-1}\in \Sigma} \eta (\xi_0, \dots, \xi_{n-1}) =1 \quad (**)

(obviously, since \eta must still be a probability). Since there are only a finite number of elements of the measure space, we can ensure first that the latter condition holds, then ensure the validity of (*) by more adjustments, making sure any adjustments upwards are balanced by adjustments downwards to guarantee (**) This brings into question whether there will not be one of these adjustments that will turn the measure of one of these elements negative. But since we’re only dealing with a finite number of points of the measure space, and our adjustments are arbitrarily small, we must just make sure that none of the adjustments are as big as the least measure of any of the elements. If there are any elements of zero measure, we adjust the measure \lambda to avoid that, whilst still remaining close to $\latex \nu$.

Now pick a longest sequence \xi^0, \xi^1, \dots ,\xi^{r-1} of elements of \Sigma^n such that

  1. (\xi^{i}_{1}, \dots ,\xi^{i}_{n-1}) = (\xi^{i+1}_{0},\dots ,\xi^{i+1}_{n-2} for any i=0,1, \dots ,r-1 and
  2. |\{ i: 0\leq i < r, \xi^i = \xi \}| \leq N\eta(\xi) for any \xi \in \Sigma^n,

where \xi^i = (\xi^{i}_{0}, \dots , \xi^{i}_{n-1}) and \xi^r = \xi^0.

The first condition clearly has to do with the translation operator, and requires that the i-th element of \Sigma^n is the translate of the (i+1)-th element, with the last term of the element not dependent on the next element. Notice that there is nothing in the first condition which prevents “looping” the same sequence over and over again. If such were allowed, we would not be able to speak of the longest sequence, and so the second condition is introduced in order to limit the lengths.

Now, suppose that equality does not hold in (2), for some \xi, remembering that N\eta (\xi) is an integer, and also supposing that \eta (\xi) \neq 0. In that case, we will be able to extend the sequence to increase the term on the left-hand side (by using closed loops), contradicting the assumption of the longest sequence. Thus, we can conclude that equality holds in (2) for each \xi.

We define a periodic element \beta of [0,1]^{\mathbb{N}} with period n+r-1 by

\displaystyle (\beta(0), \beta(1), \dots ,\beta(n+r-2)) = (\xi_{0}^{0}, \xi_{1}^{0}, \dots, \xi_{n-1}^{0}, \xi_{n-1}^{1}, \xi_{n-1}^{2},\dots ,\xi_{n-1}^{r-1})

Thus, we have found a periodic element in the neighbourhood U of \nu for each U. Taking neighbourhoods U_n of \nu such that \cap_{n} U_n = \nu, denote each corresponding periodic element by \alpha_n. For t_1, t_2, \dots a sequence of positive integers increasing sufficiently fast, set

\displaystyle \alpha(n) = \alpha_m (n-T_m)

for any n\in \mathbb{N} with T_m \leq n \leq T_{m+1}, where T_0 = 0 and T_i = T_{i-1} +c_i t_i (recall that c_n denotes the period of \alpha_n). This gives us our desired typical element, and completes the argument.


Kamae, T., A simple proof of the ergodic theorem using nonstandard analysis, Israel Journal of Mathematics, Vol. 42, No. 4, 1982

Focus, focus, focus

This is the sine qua non of mathematics. Yet even though many books have been written about this, and I have been doing maths for many years, it is still an elusive beast. My younger self would have agreed to its importance, but for some reason I have spent years trying to get by in my chosen profession without really developing this fundamental skill.

Our society values focus without encouraging it. Much is made of attention deficits in children, but how many adults work in surroundings that require absolute focus? We require children to sit quietly and pay attention in boring classes all day, but very few jobs would require that of an adult. I love learning, and I certainly enjoyed some classes at school, but most of the time I was bored out of my mind and kept quiet because I was afraid of getting low marks (which would ruin one’s life, I was told) or of getting sent to the principal or one of his minions, with the spectre of a caning always looming.  However, if you look at how so many office workers conduct their day, it is replete with interruptions, e-mails, Facebook posts, meetings, phone conversations, etc. How many professional people can go an hour without checking either their phone or their e-mail?

I would go so far as to say that many jobs actively discourage focus. The grown-up version of not doing your homework is not replying to your boss’s e-mail quickly enough, or missing a call from a client.

A lot of what I do is stultifying – student admissions, tuition meetings and administrative matters which have me squirming in my seat (and checking my e-mail). I am fortunate enough though that most of my job involves things I like. Working out second-year calculus problems may sometimes be a little frustrating if I want to get to research, but fundamentally I enjoy doing maths problems and it is often a relief to do something easy. Also, I have to admit, the boring, less challenging stuff has its own attraction. It is easier to have a discussion on the syllabus of some course when you are well and truly stuck on a problem; you at least feel that you have earned part of your keep.

And that is the trap, one I must confess to having fallen into. Bureaucracy increases exponentially – the more bureaucrats there are, the more bureaucracy they generate. As such, I am never without some task awaiting my attention or ten e-mails waiting to be answered. Most of these tasks do not take a lot of time. They can usually be done in less than half an hour, and you have that slight feeling of accomplishment at having something you can cross off your list. I think that continually facing such things has attenuated my mathematical attention. I cannot remember the last time I have sat for hours on a single paper or book, completely engrossed in trying to understand it. I do maths here and there, correcting this paper, moving on to clearing my inbox, then trying to read a new theorem, remembering something urgent about an exam paper that I have to discuss with a colleague, and so on. When your mind operates thus for a while, it becomes ingrained, and it gets uncomfortable when you focus on one thing for too long.

Clearly, once recognised this problem could not be allowed to persist. Lines had to be drawn in the sand. The very first thing I did was to dedicate the first hour of my working day to reading mathematics with the fullest concentration I could muster. I have previously mentioned the book on nonstandard number theory I am reading, and decided to make that my focus.  Unless summoned to a meeting I would do this directly after arriving at work and clocking in (literally – at our university you have to). I set aside a desk in my office which has nothing on it but some scratch paper. This part is very important – when I work at my desk with my computer, I will inevitable glance up at it from time to time, and that breaks the flow. Next, I put on wireless headphones with very calm music, something which manages to block external sounds but is not itself interesting enough to distract you. I sometimes use the sounds of ocean waves or rain falling, to the same effect. For that one hour, I don’t have to check my phone and don’t get up at all.

The first realisation from this practice was that concentration is its own reward. Giving myself the mental space to devote solely to mathematics not only reminded me of why I do it, but the very act of being intensely focused is extremely pleasant. I’m not saying it is always particularly easy, but it is worth it. Soon after starting to do this every morning, I found myself craving it. The clarity of mind during that hour is wonderfully refreshing. You have one thing to do, and only that. And when you get to the rest of the day, with its e-mails and interruptions, you have already learned something valuable, and done it well.

This worked so well that I found myself sneaking in extra hours of concentrated work whenever I could manage. It should be mentioned that this does not work as well with shorter periods of time. Thirty minutes is still not useless, but because the end is in sight, I find that I do not immerse myself as deeply. I often find myself longing to go on at the end of my hour, but do not push it to the point of fatigue. If I still feel eager to go on, I am more likely to create more time for it later. It also helps to be in the middle of an interesting proof, so I can’t wait to get back to it.

Cultivating mathematical attention is not as simple as this, though. Reading and research are two different animals. Both are creative activities, since learning anything well cannot be done passively. But the same method does not quite work for doing original mathematics, and this I consider the difficult bit. Since this deserves an entire post of its own, I will continue this later (after a maths-related post, of course).

The Laws of Meetings

  1. They who schedule many meetings have obscenely little of value to do. They who enjoy meetings have nothing of value to do.
  2. When you feel you have to schedule a meeting, you have already failed. Go into it with this knowledge, and atone.
  3. If you schedule a meeting to accomplish anything that could have been done with free project software or otherwise accomplished online, you have failed doubly.
  4. Capacity for productive work falls exponentially during a meeting, but only increases logarithmically afterwards.
  5. If you have called the meeting and speak more than all others attending combined, it is not a meeting.
  6. It is okay to go to a meeting just for the snacks.