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For a measure space $(X,\mathcal{A},\mu)$, the space of "observables" with respect to finite set $F$ which is endowed with counting measure on all of its subsets, is defined as follows:

$$obs (X, \mu, F)=\{f:X\to F \,|\, f \,\,\text{is measurable}\}.$$

$obs(X, \mu, F)$ With metric $d$ which measures the points of difference of two functions is a metric space:

$$d(f, g)=\mu \{x|f(x) \neq g(x)\}. $$

My questions are:

1- where the term "observable" comes from and why?

2- Which connections may I expect from this correspondence, by correspondence I mean corresponding a metric space $obs(X, \mu, F)$ to a measure space $(X,\mathcal{A},\mu)$, and which connection exists?

PS:

  • For clearing the question 2, I must explain some more details. $(obs(X, \mu, F), d)$ is a compact metric space (It is compact because pointwise limit of measurable functions is measurable). Now my new question is:
    2': When I must expect for observables space be a geodesic metric space and which geometric properties of this space relates to measure theoretic properties of the space $(X,\mathcal{A},\mu)$?

  • As @michael-greinecker 's mentioned, considering the finite set $F$ with counting measure is not important for measurability of observables.

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    $\begingroup$ Question 1: if you think of $X$ as the set of states of a physical system, or the set of possible outcomes of an experiment, then any function on $X$ is an "observable" quantity that returns different values depending on the state the system is in. Question 2: unclear what you are asking. $\endgroup$
    – Nik Weaver
    Commented Nov 12, 2017 at 16:26
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    $\begingroup$ These observables are nothing else than what is commonly called "finite measurable partitions" - which are widely used in ergodic theory. $\endgroup$
    – R W
    Commented Nov 12, 2017 at 18:04
  • $\begingroup$ @NikWeaver you are right, my second question is unclear, but I don't mean any functoriality because if it is then it is a poor one. This correspondence appears in ergodic theory specially in definition of entropy. I must add a PS to clear my question. $\endgroup$
    – Shakiba
    Commented Nov 12, 2017 at 18:25
  • $\begingroup$ @RW Yes, but I expect a physical interpretation which Nik explained somewhat. $\endgroup$
    – Shakiba
    Commented Nov 12, 2017 at 18:59
  • $\begingroup$ Where does counting measure on subsets of $F$ come in? $\endgroup$
    – Michael Greinecker
    Commented Nov 12, 2017 at 20:09

1 Answer 1

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The space of observables is typically not compact when $|F| > 1$. For example, if $|F| = 2$ and $X$ is a nonatomic probability space then the space of observables can be identified with the measure algebra of $X$.

Futhermore, any two nonatomic standard probability spaces are isomorphic and hence under this assumption on $X$ the space of observables depends only on the cardinality of $F$.

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  • $\begingroup$ For any set $X$ and finite set $F$, the topological space $F^{X}$ with product topology which is pointwise convergence, is a compact space (Tychonoff's theorem). But measurable elements if this space is a closed subset so is compact. $\endgroup$
    – Shakiba
    Commented Nov 13, 2017 at 17:49
  • $\begingroup$ @Shakiba, burtonpeterj is right. (The set of measurable elements is typically not closed.) $\endgroup$
    – Nik Weaver
    Commented Nov 13, 2017 at 18:15
  • $\begingroup$ @nik-weaver For finiteness of $F$, any convergent sequence of measurable, $f_{n}$, which is convergent to a function $f$, when is computed in a point must be ultimately constant so there is a &n& such that $f_{n}=f$, thus $f$ is measurable. Am I wrong? $\endgroup$
    – Shakiba
    Commented Nov 13, 2017 at 18:47
  • $\begingroup$ @nik-weaver oops! my reasoning was incorrect. Let me try a different reasoning: measurability of $f$ means $f^{-1} (a)$ is measurable for any $a \in F$ but when a sequence of measurable functions $f_{n}$ converges to $f$, $f^{-1}(a)=\{x \in X | \exists N, \forall n \geq N, x \in f_{n}^{-1}(a)\}$. Latest equality is nothing but $\bigcup_{k=1} \bigcap_{n=k}f_{n}^{-1}(a)$ which is measurable. So this is a proof for: pointwise limit of measurable observable is a measurable one. $\endgroup$
    – Shakiba
    Commented Nov 14, 2017 at 6:39
  • $\begingroup$ Actually, that isn't right. It only shows that the limit of a sequence of measurable functions is measurable. The limit of a net won't be, in general. And anyway the metric topology is different from the product topology. $\endgroup$
    – Nik Weaver
    Commented Nov 14, 2017 at 12:46

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