Uncertainty

How do we measure continuous quantities?

First, set up some sort of "triangulation" of the space in which the continuous quantity can take values. For example, a ruler is a "triangulated" line. Typically some sort of quantitative system is chosen to keep track of the cells; e.g. for a ruler we count how many ticks there are.

A measurement of a continuous quantity is simply an identification of which one of those intervals the continuous quantity lies inside. It's a declaration "this quantity lives inside this interval."

How do we apply functions to continuous quantities?

Motivating example

Let's suppose that we are measuring lengths with a meter stick, which goes down to the nearest 2 millimeters. That is, the ticks on the stick appear at 0mm, 2mm, 4mm, 6mm, etc.

Suppose that we measure a certain length, then, by some process, we double the length. That is, we apply the "continuum" function $f(x)=2x$.

That is what $f$ does to length itself, but what does $f$ do to measurements of length?

Let $x$ denote a length lying somewhere in the range $[2k-2,2k]$ millimeters (a minimal interval on the meter stick). Then $f(x)=2x$ lies somewhere in the range $[4k-4,4k]=[4k-4,4k-2]\cup [4k-2,4k]$, which is the union of two minimal intervals of the meter stick. So maybe we should say that $f$ induces a multi-valued function?

Hmm, but what would be the induced function of $g(x)=1.5x$? It would send the interval $[2k-2,2k]$ to the interval $[3k-3,3k]$, which is not necessarily a union of two minimal intervals.

I think it would be best to treat the induced function in a probabilistic manner: If the length $x$ is equally likely to be anywhere in the range $[2k-2,2k]$, then

• $2x$ it has a 1/2 chance of being in $[4k-4,4k-2]$, and a 1/2 chance of being in $[4k-2,4k]$.
• $3x$ has,
  • if $k$ is even, a 2/3 chance of being in $[3k-2,3k]$, and a 1/3 chance of being in $[3k-4,3k-2]$.
  • if $k$ is odd, a 2/3 chance of being in $[3(k+1)-6,3(k+1)-4]$, and a 1/3 chance of being in $[3(k+1)-4,3(k+1)-2]$.

The general principle is that you have some sort of probability measure on the triangulation, and then you push it forward to another one.

The "problem" with this is that the true probability distributions are "continuum," as the following example demonstrates.

Let the grid on one axis consist of intervals $[n\pi /6,(n+1)\pi /6]$, let the grid on the other axis consist of intervals $[0.1n,0.1(n+1)]$ where $n\in \mathbb {Z}$, and let the continuum function $h(x)=\cos(x)$. Then $h$ sends the interval $[0,\pi /6]$ to the interval

$$[{\sqrt {3}}/2,1]=[0.866\cdots ,1]\subset [0.8,0.9]\cup [0.9,1.0]$$

So after applying $h$ to a length in that internal, we will get a length which is in the interval $[0.9,1.0]$ with probability ${\textstyle {\frac {1}{10-5{\sqrt {3}}}}\approx 0.75}$, and in the interval $[0.8,0.9]$ with probability ${\textstyle {\frac {9-5{\sqrt {3}}}{10-5{\sqrt {3}}}}\approx 0.25}$.

General formulation

Let $f:\mathbb {R} \rightarrow \mathbb {R}$ be a continuous function (which is also a continuum function), and suppose we have a grid consisting of intervals $[a_{n},b_{n}]$ where $a_{n},b_{n}$ are real numbers such that $b_{n}=a_{n+1}$. We refer to the ordered set of these intervals as "the grid" and denote it as $\Gamma$.

Definition. A probability distribution on a set $S$ is a function $p:S\rightarrow [0,1]$, such that $p_{s}=0$ for all but finitely many $s\in S$, and $\sum _{s\in S}p_{s}=1$. Let ${\text{Prob}}(S)$ denote the set of probability distributions on $S$.

Since $f$ is continuous, it sends intervals to intervals. If $I=[a,b]$ is any interval, define its volume as ${\text{Vol}}(I):=b-a$. Then $f$ induces a function ${\widetilde {f}}:\Gamma \rightarrow {\text{Prob}}(\Gamma )$, where

$${\widetilde {f}}(n)_{m}:={\frac {{\text{Vol}}([a_{n},b_{n}]\cap f^{-1}([a_{m},b_{m}]))}{{\text{Vol}}([a_{n},b_{n}])}}$$

The interpretation of ${\widetilde {f}}$ is completely clear: If the function $f$ is applied to a quantity that was measured as being in the grid interval $[a_{n},b_{n}]$, then the probability that the resulting quantity will be measured as being in the grid interval $[a_{m},b_{m}]$ is ${\widetilde {f}}(n)_{m}$.

We can say that "a measurement of $f$ relative to the triangulation $\Gamma$" is a field of random variables, one at each triangle, where the random variable at triangle $n$ is drawn from the probability distribution ${\widetilde {f}}(n)$.

Measure-theoretic interpretation

For my purposes in this section, by a measure on $\mathbb {R}$ I mean one for which all points have measure 0. $\Gamma$ is a countable cover of $\mathbb {R}$ by measurable sets. Any measure $\mu$ on $\mathbb {R}$ restricts to a measure $\mu |_{\Gamma }$ on $\Gamma$, where ${\textstyle \mu |_{\Gamma }(S):=\mu \left(\bigcup _{s\in S}s\right)=\sum _{s\in S}\mu (s).}$ In particular, if $\mu$ is a probability measure then so is $\mu |_{\Gamma }$.

For an interval $I_{n}$ of $\Gamma$, let $\mu _{n}$ be the probability measure defined by ${\textstyle \mu _{n}(A):={\frac {{\text{Vol}}(A\cap I_{n})}{{\text{Vol}}(I_{n})}}}$. Let $f:\mathbb {R} \rightarrow \mathbb {R}$ be any measurable function. Then $f$ induces a probability measure $f_{*}\mu _{n}$ on $\mathbb {R}$, and hence a probability measure $f_{*}\mu _{n}|_{\Gamma }$ on $\Gamma$. In this language, ${\widetilde {f}}:\Gamma \rightarrow {\text{Prob}}(\Gamma )$ is simply the function sending

$$I_{n}\mapsto (f_{*}\mu _{n})|_{\Gamma }.$$

One can easily check that the two formulas for ${\widetilde {f}}$ agree.

Functoriality

The above construction generalizes vastly.

First, let $X,Y$ be any measurable spaces, and let $f:X\rightarrow Y$ be a measurable function. Let $\Gamma _{X},\Gamma _{Y}$ be triangulations of $X$ and $Y$. Then $f$ induces a map $\Gamma _{X}\rightarrow {\text{Prob}}(\Gamma _{Y})$. [TODO this isn't exactly right. I need a pre-existing "volume" measure on $X$ in order to get the map $\Gamma \rightarrow {\text{Prob}}(\Gamma )$. ]

Now, if we haven't measured a quantity recently, then it might be right to think about it as a probability distribution, rather than as having a specific value. (It does have a specific value, but we don't know what it is.) In such a case as this, we can still talk about what a function $f:X\rightarrow Y$ will do to the quantity. Indeed, if the quantity is in a specific cell of $\Gamma _{X}$, then we know what $f$ will do to it (in the generalized sense of probability distributions). And the quantity must be in some specific cell of $\Gamma _{X}$. So we should expect $f$ to induce a map ${\text{Prob}}(\Gamma _{X})\rightarrow {\text{Prob}}(\Gamma _{Y})$, which agrees with $\Gamma _{X}\rightarrow {\text{Prob}}(\Gamma _{Y})$ when restricted via $\Gamma _{X}\rightarrow {\text{Prob}}(\Gamma _{X})$.

There is a map $\alpha :{\text{Prob}}(\Gamma _{X})\rightarrow {\text{Prob}}(X)$, defined as follows:

$$\alpha (p):=\left(A\mapsto \sum _{\Delta \in \Gamma _{X}}p(\Delta ){\frac {{\text{Vol}}(A\cap \Delta )}{{\text{Vol}}(\Delta )}}\right).$$

This $\alpha$ satisfies the property that $\alpha (p)|_{\Gamma _{X}}=p$, but the composition with restriction in the other direction is most definitely not the identity. [TODO I expect that the composition in the other direction is "homotopic" to the identity, because it only differs from the original in a local way.]

So we get a map $\rho _{Y}\circ f_{*}\circ \alpha _{X}:{\text{Prob}}(\Gamma _{X})\rightarrow {\text{Prob}}(\Gamma _{Y})$ where $\rho _{Y}:{\text{Prob}}(Y)\rightarrow {\text{Prob}}(\Gamma _{Y})$ is the restriction, whenever we have an $f:X\rightarrow Y$. However, this construction is not functorial, because $\alpha \circ \rho \neq {\text{Id}}.$ Another reason it can't be functorial is that "triangulations" are not part of the initial data. Yet another reason it can't be functorial is that the identity map on $X$ can actually give interesting maps between different triangulations of $X$.

Renormalization

Let $f:\mathbb {R} \rightarrow \mathbb {R}$ be some continuum function. There is a relationship between $f$ as measured on a fine triangulation, and $f$ as measured on a coarse triangulation.

Motivating example

Let's suppose that we are measuring lengths with two meter sticks, one which goes down to the nearest 2 millimeters, and one which goes down to the nearest 1 millimeter. That is, the ticks on the first stick appear at 0mm, 2mm, 4mm, 6mm, etc., and the ticks on the second stick appear at 0mm, 1mm, 2mm, 3mm, etc. The former stick defines a triangulation which we will call $\Gamma$, and the latter defines a triangulation which we will call $\Gamma '$.

There is a (discontinuous!!) function $i:\Gamma '\rightarrow \Gamma$, sending $[0,1]\mapsto [0,2]$, $[1,2]\mapsto [0,2]$, $[2,3]\mapsto [2,4]$, $[3,4]\mapsto [2,4]$, etc. This function is the answer to the question: "If a quantity is measured in a given grid cell of the fine triangulation, which grid cell of the course triangulation will it be measured in?"

More generally, the grid cells of the fine triangulation might not be contained inside the grid cells of the coarse triangulation, so a quantity with a given fine-scale measurement might have multiple possible coarse-scale measurements. So really the answer to the italicized question above should not be a function $\Gamma '\rightarrow \Gamma$, but rather should be a function $\Gamma '\rightarrow {\text{Prob}}(\Gamma )$.