Appendix1.1 Measurability Lemma A.1

For each \(n\in \mathbb \), the \(\mathbb _d\)-valued map on \(\mathbb _d\times \mathcal ^n \times \Omega \) defined in (2.3), i.e.,

$$\begin \Gamma (\varrho , \bar,\omega ) \ := \ \rho ^_(\bar) , \end$$

is \((\mathcal (\mathbb _d)\otimes \Sigma _n\otimes \mathcal )\)-measurable for each \(n\in \mathbb \).

Proof

This follows directly due to the fact that \(\Gamma \) is a Carath odory function in the sense that for each \((\bar,\omega )\), \(\varrho \mapsto \Gamma (\varrho , \bar,\Omega )\) is continuous in \(\varrho \) and for each \(\varrho \) the map \((\bar,\omega )\mapsto \Gamma (\varrho , \bar,\Omega )\) is \(\Sigma \times \mathcal \)-measurable. The joint measurability of \(\Gamma \) follows from the fact that Carath odory functions are jointly measurable [1, Lemma 4.51]. \(\square \)

Recall that, by Tychonoff’s theorem, \(}^}\) is a compact topological space when given the product topology coming from the discrete topology on \(\mathcal \). Furthermore, \(}^}\) is a metrizable space. Explicitly, let \(\ell (\bar, \bar)\) be given by

$$\begin \ell (\bar, \bar) = \sum _} 2^ \cdot \delta _)\ne \pi _n(\bar)} \ . \end$$

Clearly, \(\ell (\bar, \bar) = 0\) if and only if \(\bar = \bar\). Moreover, because \(\delta _)\ne \pi _n(\bar)}\) satisfies the triangle inequality, \(\ell \) is a metric on \(}^}\). It is a straightforward exercise to check that the metric topology generated by \(\ell \) is product topology. Furthermore, the topology is separable, since

$$\begin \bigcup _}\bigcup _}\(a_1, \dots , a_n)\} \ . \end$$

is a base.

If we let \(\mathcal (}^})\) denote the set of probability measures on \(}^}\) and equip \(\mathcal (}^})\) with the Prokhorov metric

$$\begin \pi (\mu , \nu ) = \inf \Big \\nu (A) \le \mu (A^\varepsilon ) + \varepsilon \text A\in \mathcal (}^}) \Big \}\ , \end$$

where, for \(A\in \mathcal (}^})\), \(A^\varepsilon \) denotes the set

$$\begin A^\varepsilon = \left\ \ : \ \inf _\in A}\ell (\bar,\bar) < \epsilon \right\} \ , \end$$

then the topology on \(\mathcal (}^})\) generated by \(\pi \) is the \(\hbox ^*\) topology on \(\mathcal (}^})\) inherited by viewing \(\mathcal (}^})\) as the Banach space dual of the set \(\mathcal (}^})\) of continuous functions on \(}^}\); see [14, Apendix III, Theorem 5].

Lemma A.2

Let \(\_}\) be a disordered perfect Kraus measurement on \(\mathcal \). The map

$$\begin \begin \mathbb _d\times \Omega&\rightarrow \mathcal (}^})\\ (\varrho , \omega )&\mapsto \mathbb _ \end \end$$

is \((\mathcal (\mathbb _d)\otimes \mathcal )\)-measurable when \(\mathcal (}^})\) is given the Borel \(\sigma \)-algebra induced by the Prokhorov metric.

Proof

We have already seen that the Borel \(\sigma \)-algebra induced by the Prokhorov metric is the same as that generated by the \(\hbox ^*\) topology, so it suffices to show that, for any continuous real-valued function f on \(}^}\) (\(f\in \mathscr (}^})\)), the map

$$\begin (\varrho , \omega )\ \mapsto \ \mathbb __}(f) \end$$

is measurable.

Because \(}^}\) is a compact space, all \(f\in \mathscr \left( }^}\right) \) are bounded. We claim that any \(f\in \mathscr \left( }^}\right) \) is a uniform limit of functions of the form \(f_n\circ \pi _n\) with \(f_n\in \mathscr \left( \mathcal ^n \right) \). This could be obtained from the Stone–Weierstrass theorem, but can also be seen directly by defining \(f_n = f \circ \pi _n^\) where \(\pi _n^(a_1, \dots , a_n, a_)=(a_1, \dots , a_n, a, \dots )\), with a an arbitrary fixed element of \(\mathcal \). The dominated convergence theorem then implies that

$$\begin \mathbb __} \left( f \right) \ = \ \lim _n \mathbb __} \left( f_n \right) \ , \end$$

hence it suffices to consider f of the form \(g\circ \pi _n\), i.e. f that depends on only finitely many variables. For such f, t

$$\begin \mathbb __}(f) \ &= \ \mathbb _^_}(g)\\ \ &= \ \sum _\in \mathcal ^n} \operatorname \left( }_; \omega }^\varrho }_; \omega }^\right) g(\bar) \ , \end$$

which is clearly measurable in \((\varrho , \omega )\). \(\square \)

The following lemma shows that a re-indexing of a sequence of measurable functions by a \(\mathbb \)-valued (or \(\mathbb \cup \\)-valued) measurable map is also measurable. The proof is similar to the proof that \(X_\tau \) is measurable where \((X_t)_\) is a stochastic process and \(\tau \) is a T-valued stopping time. For a sequence of measurable maps \((X_n)_}\) and a \(\mathbb \cup \\)-value measurable function, \(\tau \), we shall call \(X_\tau \) the stopped random variable or the re-indexing of \((X_n)_}\) with respect to \(\tau \).

Lemma A.3

Let \((X_n)_}\) be a sequence of measurable maps defined on a measurable space \((X,\mathcal )\) taking values on a measurable space \((U,\mathcal )\). If \(\tau :X \rightarrow \mathbb \cup \\) is an \(\mathcal \)-measurable random map (so that \(\\) and \(\\) are elements in \(\mathcal \)), then the map \(X_\tau \) defined by

$$\begin X_\tau (x) = X_(x) & \text \tau (x) < \infty \\ y_0 & \text \tau (x) = \infty \end\right. } \end$$

for some fixed \(y_0\in Y\) is also a measurable map.

Proof

Let \(B\in \mathcal \) then we have that

$$\begin & \ = \ \tau (x)<\infty \} \sqcup \\ & \ \tau (x) =\infty \} \ . \end$$

First note that \(\ \tau (x) =\infty \} \) is either \(\\) (if \(y_0\in B\)) or \(\emptyset \) (if \(y_0\not \in B\)); in either case, it is measurable. On the other hand, we have

$$\begin \ \tau (x)<\infty \}&= \bigcup _}\ \tau (x)=n\}\\&= \bigcup _}\left( \ \cap \\right) \ . \end$$

Therefore, we have that \(X_\tau \) is indeed \((\mathcal ,\mathcal )\)-measurable. \(\square \)

1.2 Equicontinuity

In this section, we prove Proposition 3.4:

Proposition 3.4

For fixed \(\omega \in \Omega \), the family of maps \(\left\\right\} _}\) is uniformly equicontinuous.

Recall that the maps \(\Phi ^_\omega :\mathbb _d\rightarrow [0,1]\), as defined in (3.6), are given by:

$$ \Phi _\omega ^(\varrho )\ = \ \mathbb _\left[ M_^(\bar)\right] \ . $$

Proof of Proposition 3.4

It is enough to prove that each \(\Phi _\omega ^\) is K-Lipschitz, where \(K\ge 0\) is uniform in n. To simplify notation, we write \(F_n\) for \(\Phi _\omega ^\) and, for any \(\bar\in \mathcal ^n\), write \(F_}\) to denote the function

$$\begin \varrho \ \mapsto \ \operatorname \left( }_; \omega }^ \varrho }_; \omega }^\right) \operatorname \left( \left( }_; \omega }^ \cdot \varrho \right) ^2\right) \ , \end$$

so \(F_n = \sum _\in \mathcal ^n}F_}\). Note that \(F_}\ge 0\). Letting

$$\begin g_}(\varrho ) \ := \ \operatorname \left( \left( }_; \omega }^\varrho }_; \omega }^\right) ^2\right) \quad \text \quad h_}(\varrho ) \ := \ \operatorname \left( }_; \omega }^\varrho }_; \omega }^\right) \ . \end$$

it is clear that \(g_}, h_}\ge 0\) and

$$\begin F_} \ = \ \frac} }} } & \text h_}> 0\\ 0 & \text \ . \end\right. } \end$$

Since

$$\begin \frac}(\varrho ) }}(\varrho ) } \ &= \ \operatorname \left( \left( \frac}_; \omega }^\varrho }_; \omega }^}\left( }_; \omega }^\varrho }_; \omega }^\right) }\right) }_; \omega }^\varrho }_; \omega }^\right) \\&= \ \operatorname \left( \varrho ^}_; \omega }^\left( \frac}_; \omega }^\varrho }_; \omega }^}\left( }_; \omega }^\varrho }_; \omega }^\right) }\right) }_; \omega }^\varrho ^\right) \ \le \ h_}(\varrho ) \ , \end$$

whenever \(h_}(\varrho ) > 0\). Here, we use that if \(X\ge 0\) is nonzero, then \( \frac\left( X\right) }\le \frac\le I \). It follows that the functions \(F_}\) are continuous, and thus, \(F_n\) is continuous also.Footnote 4

Now fix \(\varrho , \varrho '\in \mathbb _d\) and consider the function

$$\begin \begin \varrho _:[0, 1]&\rightarrow \mathbb _d\\ \varrho _t&:= t\varrho + (1 - t)\varrho ' \ . \end \end$$

We begin by proving the map

$$\begin t\mapsto F_n(\varrho _t) \end$$

is differentiable on (0, 1). Because \(F_n = \sum _\in \mathcal ^n}F_}\), it suffices to prove that the map

$$\begin \begin G_}: [0, 1]&\rightarrow \mathbb \\ t&\ \mapsto \ F_}(\varrho _t) \ = \ \frac}(\varrho _t)}}(\varrho _t)} \end \end$$

is differentiable on (0, 1) for each \(\bar\in \mathcal ^n\). Fix \(\bar\in \mathcal ^n\). Since \(h_}(\varrho _t)\) and \(g_}(\varrho _t)\) are both clearly differentiable, \(G_}\) is seen to be differentiable whenever \(h_}(\varrho _t)\ne 0\). On the other hand, if \(h_}(\varrho _t)=0\), then we have \(h_}(\varrho ) = h_}(\varrho ') = 0\); hence, \(F_}(\varrho ) = F_}(\varrho ') = 0\) by definition, and \(G_}\equiv 0\), so differentiability is clear. Thus, we see that \(G_}\) is differentiable for \(t\in (0,1)\) and furthermore \(h_}(\varrho _t)\) is either identically zero for all \(t\in (0,1)\) or nonzero for all \(t\in (0,1)\).

Now, for any \(t\in (0,1)\), we have

$$\begin F_n'(\varrho _t) \ = \ \sum _\in C}G'_}(t) \end$$

where

$$\begin C \ = \ \left\\,\,:\,\, h_}(\varrho _t) > 0 \text t\in (0,1) \right\} \ . \end$$

For \(\bar\in C\), we compute

$$\begin \left| G'_}(t) \right| \ &\le \ \left| \frac}(\varrho _t)}}(\varrho _t)} \right| + \left| \frac}(\varrho _t) h'_}(\varrho _t)}}(\varrho _t)^2} \right| \\&= \ \left| \frac\left( }_;\omega }^ \varrho _t }_;\omega }^ }_;\omega }^ (\varrho - \varrho ') }_;\omega }^ \right) }}(\varrho _t)} \right| \\ &\quad + \left| \frac}(\varrho _t) }}(\varrho _t)^2 } \operatorname \left( }_;\omega }^ (\varrho - \varrho ') }_;\omega }^\right) \right| \\&\le 2 \operatorname \left( \left( \frac}_;\omega }^ \varrho _t }_;\omega }^}\left( }_;\omega }^ \varrho _t }_;\omega }^\right) }\right) }_;\omega }^ |\varrho - \varrho '| }_;\omega }^ \right) \\&\quad + \operatorname \left( \left( \frac}_;\omega }^ \varrho _t }_;\omega }^ }\left( }_;\omega }^ \varrho _t }_;\omega }^\right) } \right) ^2 \right) \operatorname \left( }_;\omega }^ |\varrho - \varrho '| }_;\omega }^\right) \\&\le \ (2 + d) \operatorname \left( }_;\omega }^ |\varrho - \varrho '| }_;\omega }^\right) \ . \end$$

Thus,

$$\begin \left| F_n'(t)\right| \ &\le \ (2 + d)\sum _\in \mathcal ^n} \operatorname \left( }_;\omega }^\left| \varrho - \varrho '\right| }_;\omega }^\right) \nonumber \\ \ &= \ (2 + d)\operatorname \left| \varrho - \varrho '\right| \end$$

for any \(t\in (0, 1)\). By the mean value theorem, we conclude that

$$\begin \left| F_n(\varrho ) - F_n(\varrho ')\right| \ \le \ (2 + d)\operatorname \left| \varrho - \varrho '\right| \ . \end$$

Because this bound is independent of \(\varrho , \varrho '\), we see that \(F_n\) is \((2 + d)\)-Lipschitz continuous uniformly in n, which implies the desired uniform equicontinuity. \(\square \)