Is the "call" or "evaluation" function measurable?

Question

The call function is defined as

$$ \text{call}: \begin{cases} (\mathbb{R}^{I}\times I) \to \mathbb{R} \\ (f,x) \mapsto f(x) \end{cases} $$

is "$\text{call}$" a measurable function? In other words: for a random field $Z:\mathbb{R}^n\to\mathbb{R}$ and a random location $X\in\mathbb{R}^n$, is $Z(X)$ a random variable?

To address a comment: This is not simply a question of chaining measurable functions, as both $Z$ and $X$ depend on $\omega$. I.e we really have $Z(\omega):\mathbb{R}^n\to\mathbb{R}$, so $\omega\mapsto Z(\omega)(X(\omega))=\text{call}(Z(\omega), X(\omega))$ needs to be measurable.

I would guess this should be correct tough. At least for continuous random fields. But I am not sure what to search for. In a programming context "call" would be appropriate (cf. https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Function/call). I am mostly looking for a reference to cite.

Answer 1

EDIT: previously this presented a dead end, but with @FShrike's answer this does not make any sense anymore. Instead I want to elaborate on @FShrike's answer. Partly to explain it to myself.

The Continuity Approach

The main idea is: A common sufficient criterion for measurability with regard to the Borel sigma algebra is continuity. So if we get continuity, we get Borel measurability for free. This opens three questions:

1. Is our evaluation/call function actually continuous?
2. Is the Borel sigma algebra what we want?
   • is the sigma algebra we want a Borel sigma algebra at all?
   • have we generated the Borel sigma algebra from the correct topology?
3. If the evaluation/call function is not continuous, can it still be measurable?

1 Is the evaluation function continuous?

This is the main concern of @FShrike's answer. The first question is of course, what topology can we give $\mathbb{R}^I$?

It appears there is the concept of an Exponential Topology, which exists for very general $I$ (locally compact Hausdorff is sufficient). And the evaluation function is always continuous with regard to this topology.

2 Is the Borel Sigma algebra what we want?

@FShrike also addresses the question whether we have the correct topology somewhat. It appears that

the exponential topology is identical to the compact-open topology if and only if the evaluation function is continuous with regard to the compact open topology.

The compact-open topology is only defined on the set of continuous functions $C(I)$ though. But according to wikipedia, the evaluation function is continuous with regards to the compact-open topology. This implies that the two topologies do coincide, which makes the exponential topology a reasonable extension for non-continuous functions.

What sigma algebra do we want?

Let us now start from the probability side: What do we want? Due to Kolmogorov's extension theorem, probabilists like to work with the sigma algebra $\mathcal{A}$ generated by finite cylinder sets.

What does that mean?

For a finite Subset $J=\{t_1,\dots,t_n\}\subset I$ we define the finite dimensional projection

$$ \pi_J(f):= (f(t_1),\dots,f(t_n)) $$

A cylinder is of the form $\pi_J^{-1}(A_1\times\dots\times A_n)$, because for any value $t\not\in J$ the functions in this set can take any value. So informally we have

$$ \pi_J^{-1}(A_1\times\dots\times A_n) = A_1\times\dots\times A_n \times \mathbb{R}\times\mathbb{R}\times... \subset \mathbb{R}^I $$

So the sigma algebra we want is $$ \begin{aligned} \mathcal{A} &= \{\pi_J^{-1}(A), J\subset I \text{ countable}, A\in \mathcal{B}(\mathbb{R}^{|J|})\}\\ &= \sigma( \pi_J^{-1}(A), J\subset I \text{ finite}, A\in \mathcal{B}(\mathbb{R}^{|J|}))\\ &= \sigma( \pi_{\{x\}}^{-1}(A), x\in I, A\in\mathcal{B}(\mathbb{R})) \end{aligned} $$

Where the middle is ususally the definition because it meshes well with Kolmogorov's extension theorem which uses a consistent set of finite-dimensional marginal distributions.

If you view a sigma algebra as the set of questions you are allowed to ask, the sigma algebra $\mathcal{A}$ essentially allows you to ask questions about a countable set of points of your function. For general functions this does not allow you to separate a lot of functions. The set $$ \{ f\in \mathbb{R}^I : f\equiv c \} $$ is not measurable for example. But if we restrict ourselves to continuous functions, this set suddenly becomes measurable. Because we only need to know our function in the countable points in $D(I)$ for a separable $I$ due to continuity. More precisely, we have

$$ \mathcal{A}|_{C(I)} = \mathcal{B}(C(I))\tag{*} $$

where the topology of the Borel sigma algebra $\mathcal{B}(C(I))$ is generated by the metric

$$ d(f,g) := \sum_{k=1}^\infty \frac{\max\{d_k(f,g),1\}}{2^k}, \quad d_k(f,g) := \sup_{x\in\overline{B_k(0)}}|f(x)-g(x)| $$

where we assume that the ball $\overline{B_k(0)}$ with radius $k$ is compact (need Heine-Borel).

Proof of (*):

$\subseteq$:

It is sufficient to prove that $\pi_{\{x\}}$ are continuous and therefore measurable, because $\mathcal{A}$ is the smallest sigma algebra to make them all measurable.

$\supseteq$:

Because the open balls around rational polynomials form a countable basis of the topology of $C(I)$ induced by $d$, it is sufficient to prove that the open balls are in $\mathcal{A}$. If we show that $$ h_0: f\mapsto d(f_0, f) $$ is measurable with regard to $\mathcal{A}$ we are done, because then $$ B_\epsilon(f_0) = h_0^{-1}([0,\epsilon)) \in \mathcal{A} $$

To prove $h_0$ is measurable it is sufficient to show that $h_k: f\mapsto d_k(f_0, f)$ is measurable.

But because

$$ h_k(f) = \sup_{x\in \overline{B_k(0)}\cap D(I)} |f(x)-f_0(x)| = \sup_{x\in \overline{B_k(0)}\cap D(I)} |\pi_{\{x\}}(f)-\pi_{\{x\}}(f_0)| $$

is just a countable supremum of measurable functions, it is measurable. Where we have used the separability of $I$ and the continuity of $f$ and $f_0$ when intersecting with the countable dense set $D(I)$.

Comparison with the Borel(Compact Open Topology)

The compact-open topology is defined as

$$ \tau( U_K(O): K\subset I \text{ compact}, O\in \tau_{\mathbb{R}}) $$ where $\tau_{\mathbb{R}}$ are the open sets in $\mathbb{R}$ and we define $$ U_K(O):= \{ f\in C(I) : f(K)\subset O\} =\pi_K^{-1}(O\times\dots\times O). $$ The second equality is not quite rigorous because for a compact $K$ we need the carthesian product of $|K|$ many $O$, which is generally uncountable.

For continuity of the evaluation function with regard to $d$ we only need to show, that the topology generated by $d$ is larger than the compact-open topology. For this, it is sufficient to show that every $U_K(O)$ is open with regards to $d$.

So let $U_K(O)$ be given. To show it is open, we need to find an epsilon ball around any element $f\in U_K(O)$, which we now assume as given. Using Heine-Borel again, there exists some $k\in\mathbb{N}$ such that $K\subseteq \overline{B_k(0)}$

Then for $d(f,g)\le 2^{-k}$, we have $$ \sup_{x\in K} | f(x) - g(x)| \le d_k(f,g) \overset{d(f,g)\le 2^{-k}}\le 2^k d(f,g) $$ So with an appropriate $\epsilon>0$ we can ensure by $d(f,g)\le\epsilon$ that $$ \sup_{x\in K} | f(x) - g(x)|\le \eta. \tag{1} $$ Let $\eta$ be the minimum distance of $f(y)$ and $O^\complement$. This is non-zero because $f(K)$ is compact and a subset of $O$. So with (1), we can guarantee $g(K)\subset O$ for all $d(f,g)<\epsilon$. We have found our $\epsilon$ ball.

Comparison with Borel(Exponential Topology)

The borel sigma algebra of the exponential topology would have to be smaller than the one generated by all finite projections. Given that

I am reasonably sure that the projections $\pi_J$ are measurable in the Borel-algebra, but I doubt that the Borel-algebra equals the cylinder algebra. I don't know for sure. To me, that is quite a weird thing to do, to take a product algebra for $I=\Bbb R^n$ (which is enormous!) - @FShrike

If the $\pi_J$ are measurable in the Borel-algebra, the smallest algebra to make them measurable $\mathcal{A}$ is going to be smaller or equal. If they are not equal, the measurability with regard to the Borel-algebra does not imply measurability with regard to $\mathcal{A}$.

Because I do not really understand the definition of the Exponential Topology well in the first place, I doubt that I will be able to make a more concrete statement in the near future.

Answer 2

This post assumes $\Bbb R^I$ is taken with the compact-open topology, and is given the Borel sigma-algebra. $\Bbb R^I$ is used to denote the set of continuous functions $I\to\Bbb R$, so only addresses a special subcase of the problem.

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Reference.

You mean the domain to be $\Bbb R^I\times I$, by the way.

The answer is yes since $I$ is locally compact Hausdorff in the sense that it is Hausdorff and there is a compact local base at every point. That means the evaluation is continuous, so in particular it is Borel measurable.

That is one condition for the exponential topology on $X^I$ (for any other space $X$) to exist. In fact, the compact-open topology on $V^U$ is exponential iff. the evaluation (‘call’): $$V^U\times U\overset{\rm{ev}}{\longrightarrow}V$$Is continuous.

By an ‘exponential topology’, I mean a topology on the set $V^U$ such that the adjunction pairing: $$f\mapsto\mathrm{ev}\circ(f\times1):W\times U\to V$$Among continuous $f:W\to V^U$ defines a bijection $C(W,V^U)\cong C(W\times U,V)$ for all spaces $W$. Under certain conditions, such as when one works in the category of compactly generated Hausdorff spaces, that last bijection is even a homeomorphism!

Note: the book I reference does dip a little into compactly generated Hausdorff spaces, but it doesn’t do so very well or in great detail / rigour. Also, be warned that conflicting definitions of ‘compactly generated’ can cause serious problems! There is a paper by Strickland, if I remember correctly, that covers compactly generated spaces and their nice categorical properties in good detail.