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?

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 ${\textstyle {\frac {9-5{\sqrt {3}}}{10-5{\sqrt {3}}}}\approx 0.25}$.