Invariant measures for place-dependent idempotent iterated function systems

Abstract

We study the set of invariant idempotent probabilities for place-dependent idempotent iterated function systems defined in compact metric spaces. Using well-known ideas from dynamical systems, such as the Mañé potential and the Aubry set, we provide a complete characterization of the densities of such idempotent probabilities. As an application, we provide an alternative formula for the attractor of a class of fuzzy iterated function systems.

Appendix A: Fundamentals of idempotent analysis

The following exposition is based on the ideas of Kolokol’tsov and Maslov [16], who considered a different setting for applications of idempotent analysis. See also Refs. [5, 9, 10, 17, 18, 21, 23, 32, 33], among many others for additional references.

1.1 Proof of Theorem 1.2

The proof of Theorem 1.2 will be divided in several parts as exposed below.

Proposition 6.1

Any $m\in C^*(X,\mathbb {R})$ is order preserving, that is, if $f \le g$ then $m( f) \le m( g)$. Moreover

$$\begin{aligned} \min _{X} f \le m(f) - m(0) \le \max _{X} f. \end{aligned}$$

Proof

If $f \le g$ then $f \oplus g = g$. As m is max-plus additive, we obtain $m(g) = m(f \oplus g)=m( f) \oplus m( g)$, thus $m( f) \le m( g)$.

As we have a function $m: C(X,\mathbb {R}) \rightarrow \mathbb {R}$, then we get $m(0)\ne -\infty$. As for any $f \in C(X, \mathbb {R})$, $\min _{X} f \le f(x) \le \max _{X} f$, we obtain

$$\begin{aligned}{} & {} m(\min _{X} f ) \le m(f) \le m(\max _{X} f) \\{} & {} m((\min _{X} f)\odot 0) \le m(f) \le m((\max _{X} f)\odot 0) \\{} & {} (\min _{X} f) \odot m(0) \le m(f) \le (\max _{X} f) \odot m( 0). \end{aligned}$$

$\square$

The semimodule $\mathcal {V}:=(C(X,\mathbb {R}), \oplus , \odot )$ can be topologized with the usual structure given by the metric $d_{\infty }(f,g)=\bigoplus _{x\in X} |f(x)- g(x)|$. Introducing a metric $\rho : \mathbb {R}_{\max } \times \mathbb {R}_{\max } \rightarrow [0, +\infty )$ given by $\rho (a,b):=|\exp (a)-\exp (b)|$, we obtain a topological semiring $(\mathbb {R}_{\max }, \rho )$ and for the set $\mathcal {F}(X,\mathbb {R}_{\max }):=\{ f: X \rightarrow \mathbb {R}_{\max }\}$, we can consider the topology given by

$$\begin{aligned} d_{\rho }(f,g)=\bigoplus _{x\in X} \rho (f(x), g(x))=\bigoplus _{x\in X} |\exp (f(x))- \exp (g(x))|, \end{aligned}$$

for any $f, g \in \mathcal {F}(X,\mathbb {R}_{\max })$.

Proposition 6.2

Any $m\in C^*(X,\mathbb {R})$ is nonexpansive with respect to the usual sup-norm $d_{\infty }$ in $C(X, \mathbb {R})$ and the absolute value $|\cdot |$ in $\mathbb {R}$. In particular, it is continuous.

Proof

Indeed,

$$\begin{aligned}&\displaystyle -d_{\infty }(f, g) + g(x) \le f(x) \le d_{\infty }(f, g) + g(x), \forall x \in X\\&\displaystyle -d_{\infty }(f, g) + m(g) \le m(f) \le d_{\infty }(f, g) + m(g)\\&\displaystyle |m(f) - m(g)| \le d_{\infty }(f, g). \end{aligned}$$

$\square$

Definition 6.3

We say that a sequence of functions $f_n \in C(X,\mathbb {R})$ converges pointwise to a function $f \in \mathcal {F}(X,\mathbb {R}_{\max })$ if $\lim _{n\rightarrow \infty } \rho (f_n(x),f(x)) =0$, for any $x\in X$. Equivalently:

• 1. $\lim _{n\rightarrow +\infty }f_n(x) = f(x)$, for all $x\in X$ such that $f(x)\in \mathbb {R}$

• 2. $\lim _{n\rightarrow +\infty }f_n(x) = -\infty$, for all $x\in X$ such that $f(x)=-\infty .$

Definition 6.4

Given $a \in \mathbb {R}$ and $x \in X$, a fixed point, we define the Dirac function

$$\begin{aligned} g^{a}_{x}(y):= \left\{ \begin{array}{ll} a, &{}\quad y=x \\ -\infty , &{}\quad y\ne x \end{array} \right. \end{aligned}$$

and $\Delta (X, \mathbb {R})$ the set of these functions.

Lemma 6.5

There exists a monotone nonincreasing sequence of continuous functions $f_n\in C(X,\mathbb {R})$ which converges pointwise to $g^{a}_{x}$.

Proof

A consequence of the fact that X is compact metric space is that we can easily build a sequence in $C(X, \mathbb {R})$ of nonincreasing functions converging pointwise to $g^{a}_{x}$. Indeed, for each $n \in \mathbb {N}$, consider the function $f_{n}$ defined by

$$\begin{aligned} f_{n}(y):= \left\{ \begin{array}{ll} a, &{}\quad d(y,x)<1/n \\ -n, &{}\quad d(y,x)>2/n \\ a-(n^2 + a n)(d(y,x) - 1/n), &{}\quad 1/n \le d(y,x) \le 2/n \end{array} \right. . \end{aligned}$$

(16)

It is obviously continuous in y and monotone nonincreasing in n. Moreover, for any $y \in X$, $y\ne x$, we get $d(y,x)>2/n$ for n large enough. In this case, we obtain $f_n(y)\rightarrow -\infty$. This proves the pointwise convergence. $\square$

Definition 6.6

Given $g^{a}_{x}$, we will call the sequence of functions $(f_{n})$ given in (16) as the standard continuous approximation of $g^{a}_{x}$.

Lemma 6.7

Let $(f_n)$ be a nonincreasing sequence of continuous functions converging pointwise to a continuous function f in a compact set X. Then the convergence is uniform.

Proof

Suppose in contradiction there is an $\varepsilon >0$ and a sequence of points $(x_n)\in X$ such that $f_n(x_n)>f(x_n)+\varepsilon ,\; \forall \,n\in \mathbb {N}$. As X is compact, there exists a subsequence $(x_{n_i})$ and a point $x_0\in X$ such that $x_{n_i}\rightarrow x_0$. Let $n_j$ be such that $f_{n_j}(x_0)<f(x_0)+\varepsilon /2$ (it exists because we have convergence pointwise). As f and $f_{n_j}$ are continuous in $x_0$, there exists a $\delta >0$ such that $d(x,x_0)<\delta \Rightarrow f_{n_j}(x)<f(x)+\varepsilon$. As $(f_n)$ is nonincreasing, we also get

$$\begin{aligned} (d(x,x_0)<\delta ,\,n\ge n_j) \Rightarrow f_{n}(x)<f(x)+\varepsilon . \end{aligned}$$

This is a contradiction because, by definition of $x_n$ and $x_0$, for n large enough, we must have $d(x_n,x_0)<\delta$ and also $f_n(x_n)>f(x_n)+\varepsilon$. $\square$

Lemma 6.8

Let $\bar{C}(X,\mathbb {R})$ be the set of functions $g:X\rightarrow \mathbb {R}_{\max }$ such that there exists a nonincreasing sequence of continuous functions $(f_n)$ converging pointwise to g. This set is closed with respect to $(\oplus ,\odot )$ operations. Given $m \in C^{*}(X, \mathbb {R})$ an idempotent measure, we can extend it to an idempotent measure on $\bar{C}(X,\mathbb {R})$ by $\tilde{m}(g) = \lim _{n\rightarrow +\infty }m(f_n)$, where $(f_n)$ is any nonincreasing sequence of continuous functions converging pointwise to g.

Furthermore, if C is a set which is closed with respect to $(\oplus ,\odot )$ operations, such that $C(X,\mathbb {R})\subseteq C \subseteq \bar{C}(X,\mathbb {R})$ and $\overline{m}$ is an idempotent measure on C which extends m, then $\overline{m}\le \tilde{m}$.

Proof

If $g,g'\in \bar{C}(X,\mathbb {R})$ and $f_n\rightarrow g,\,f'_n\rightarrow g'$, we have $f_n \oplus f'_n \rightarrow g\oplus g'$ and $a\odot f_n \rightarrow a\odot g$ then, $\bar{C}(X,\mathbb {R})$ is closed with respect to $(\oplus ,\odot )$ operations.

Consider any sequence of monotonous nonincreasing continuous functions $(f_{n})$ converging pointwise to g. We claim that the value $\tilde{m}(g):=\lim _{n \rightarrow \infty } m(f_{n}) \in \mathbb {R}_{\max }$ is well defined and furthermore it does not depend on $(f_n)$.

Indeed, as $f_{n} \ge f_{n+1}$, we have $m(f_{n}) \ge m(f_{n+1})$. Thus, there exists the limit $\lim _{n \rightarrow \infty } m(f_{n}) \in \mathbb {R}_{\max }$. Let $(f_{n})$ and $(f'_{n})$ be sequences of monotonous nonincreasing continuous functions converging pointwise to g. For any fixed k, we define a new sequence $\psi _{n}:=f_{k}\oplus f'_{n},$ so that $\psi _{n}$ converges pointwise to $f_{k} \in C(X,\mathbb {R})$. Applying Lemma 6.8, we get that $\psi _n$ converges uniformly to $f_k$. From Proposition 6.2, we obtain $m(f_k) = \lim _{n \rightarrow \infty } m(\psi _{n})$. Then $m(f_k)= \lim _{n \rightarrow \infty } m(f_k \oplus f'_{n}) \ge \lim _{n \rightarrow \infty } m(f'_{n}).$ Now we can take the limit on the left hand side obtaining $\lim _{k\rightarrow \infty }m(f_k) \ge \lim _{n \rightarrow \infty } m(f'_{n}).$ Reversing the role of the sequences, we obtain that the limits are equal.

It is easy to see that $\tilde{m}$ is actually an extension of m because, for $h \in C(X, \mathbb {R})$, we can take the sequence $f_{n}=h, \forall n$, which is continuous, monotone, and not increasing, so $\tilde{m}(h)= \lim _{n \rightarrow \infty } m(h)= m(h)$.

Now we prove that $\tilde{m}$ is idempotent. If $g,g'\in \bar{C}(X,\mathbb {R})$ and $f_n\rightarrow g,\,f'_n\rightarrow g'$, we have

$$\begin{aligned} \tilde{m}(g\oplus g')= & {} \lim _{n \rightarrow \infty } m(f_n\oplus f'_n) = \lim _{n \rightarrow \infty } [m(f_n)\oplus m(f'_n)] =\\= & {} [\lim _{n \rightarrow \infty } m(f_n)]\oplus [\lim _{n \rightarrow \infty } m(f'_n)]=\tilde{m}(g)\oplus \tilde{m}(g'). \end{aligned}$$

and $\tilde{m}(a\odot g) = \lim _{n \rightarrow \infty } m(a\odot f_n) = \lim _{n \rightarrow \infty } [a\odot m(f_n)] = a\odot [\lim _{n \rightarrow \infty } m(f_n)]=a\odot \tilde{m}(g).$

Finally, if C is a set closed with respect to $(\oplus ,\odot )$ operations, such that $C(X,\mathbb {R})\subseteq C \subseteq \bar{C}(X,\mathbb {R})$ and $\overline{m}$ is an idempotent measure on C which extends m, then given $g\in C$ and any nonincreasing sequence of continuous function $(f_n)$ converging pointwise to g we have $g\le f_n$. Therefore, $\overline{m}(g) \le \overline{m}(f_n) = m(f_n )$ and taking the limit in n, we get $\overline{m}(g) \le \tilde{m}(g)$. $\square$

Let us remember the definition of u.s.c. function.

Definition 6.9

A function $f: X \rightarrow \mathbb {R}_{\max }$ is called upper semi-continuous (u.s.c. for short) at a point $x_0 \in X$ if for every real $c> f\left( x_0\right)$, there exists a neighborhood U of $x_0$ such that $f(x)< c$ for all $x \in U$. Equivalently, f is upper semi-continuous at $x_0$ if and only if

$$\begin{aligned} \limsup _{x \rightarrow x_0} f(x) \le f\left( x_0\right) . \end{aligned}$$

Next Lemma is inspired in [16, Lemma 1].

Lemma 6.10

Given an idempotent measure $m \in C^{*}(X, \mathbb {R})$, consider the extension $\tilde{m}$ from Lemma 6.8.

• 1. The map $F:\mathbb {R} \times X \rightarrow \mathbb {R}_{\max }$ defined by $F(a,x)=\tilde{m}(g^{a}_{x})$ is u.s.c. with respect to x and monotonous in $a \in \mathbb {R}$.

• 2. $F(a,x)=a \odot F(0,x)$, for any $a \in \mathbb {R}, x \in X$;

• 3.
  $$\begin{aligned} m(h)=\bigoplus _{x \in X} F(h(x),x) = \bigoplus _{x \in X} F(0,x)\odot h(x), \end{aligned}$$

  for all $h \in C(X, \mathbb {R})$.

Proof

(1) If $a> a'$, then $g_x^a \ge g_x^{a'}$. So $F(a,x)= \tilde{m}(g_x^a) \ge \tilde{m}(g_x^{a'}) = F(a',x)$. We now prove that the correspondence $x \rightarrow F(a,x)$ is u.s.c. Fix $x_0 \in X$ and take any $c > F(a,x_0)$. Recall that $F(a,x_{0})=\tilde{m}(g^{a}_{x_{0}})= \lim _{n \rightarrow \infty } m(f_{n})< c$ for $(f_{n})$ the standard continuous approximation of $g^{a}_{x_0}$. So there is $N_{c}$ such that for all $n_{0}> N_{c}$ we have, $m(f_{n_{0}})< c$. Consider $x \in U=B_{\frac{1}{n_{0}}}(x_{0})$ and choose n big enough so that $f'_{n} \le f_{n_{0}}$ where $(f'_{n})$ is the standard continuous approximation of $g^{a}_{x}$.

Under this hypothesis, we have $m(f'_{n}) \le m(f_{n_{0}})$ and taking the limit on the left side, we get $\lim _{n \rightarrow \infty }m(f'_{n}) \le m(f_{n_{0}})$, or equivalently $F(a,x) \le m(f_{n_{0}})<c.$ This proves that the correspondence $x \rightarrow F(a,x)$ is u.s.c.

(2) $F(a,x)= \tilde{m}(g_x^a) = \tilde{m}(a\odot g_x^{0}) =a \odot \tilde{m}(g_x^{0})=a\odot F(0,x)$.

(3) By applying (2), we get the second equality and then we just need to prove that

$$\begin{aligned} m(h)=\bigoplus _{x \in X} F(h(x),x) \end{aligned}$$

for all $h \in C(X, \mathbb {R})$. As $g_x^{h(x)}(y) \le h(y),\, \forall \, x,y\in X$, we obtain $\tilde{m}(g_x^{h(x)})\le \tilde{m}(h) = m(h)\,\forall \,x\in X$. Then

$$\begin{aligned} \bigoplus _{x \in X} F(h(x),x) = \bigoplus _{x \in X} \tilde{m}(g_x^{h(x)}) \le \bigoplus _{x \in X} m(h) = m(h). \end{aligned}$$

To prove the opposite inequality, we consider any $\varepsilon >0$. Denote by $f_n^x, n\ge 0$ the standard approximation of $g_x^{h(x)}$ as given in (16). For each $n\ge 0$, let $x_n\in X$ be such that $\displaystyle {\oplus _{x\in X}m(f_n^x) < m(f_n^{x_n})+\varepsilon }$. As X is compact, there exists a point $\tilde{x}$ and a subsequence $(x_{n_i})$ such that $x_{n_i}\rightarrow \tilde{x}$. As h is continuous, there exists $k_0$ such that $h(x)<h(\tilde{x})+\varepsilon$ for any $x\in B(\tilde{x},\frac{1}{k_0}).$ For each natural $k\ge k_0$, there exists $N>k$ such that $B(x_{n_i},\frac{2}{n_i})\subset B(\tilde{x},\frac{1}{k})$ for any $n_i\ge N$. As $h(x_{n_i})<h(\tilde{x})+\varepsilon$ and $B(x_{n_i},\frac{2}{n_i})\subset B(\tilde{x},\frac{1}{k})$, it follows from definition (16) that $f_{n_i}^{x_{n_i}} \le f_{k}^{\tilde{x}}+\varepsilon$ for $n_i\ge N$ and consequently, for $n_i\ge N$, we obtain

$$\begin{aligned} \oplus _{x\in X}m(f_{n_i}^x)< m(f_{n_i}^{x_{n_i}})+\varepsilon < m(f_{k}^{\tilde{x}})+2\varepsilon . \end{aligned}$$

(17)

As h is continuous and X is compact, the function h is uniformly continuous. Then there exists $\delta$ such that $d(x,y)<\delta \Rightarrow |h(y)-h(x)|<\varepsilon$. Consider a finite cover of X by balls of radius $\frac{1}{n_i}< \delta$, with $X\subset B(z_1,\frac{1}{n_i}) \cup \cdots \cup B(z_{l_i},\frac{1}{n_i})$. If $y\in B(z_j,\frac{1}{n_i})$ then $h(y) <h(z_j)+\varepsilon$. Furthermore, as from definition (16), $f_{n_i}^{h(z_j)}(x)= h(z_j)$ for any $x\in B(z_j,\frac{1}{n_i})$ we get, $h \le \oplus _{z_j}f_{n_i}^{z_j} +\varepsilon .$ Finally, applying (17) we have,

$$\begin{aligned} m(h) \le m\left( \oplus _{z_j}f_{n_i}^{z_j} +\varepsilon \right) = \oplus _{z_j}m(f_{n_i}^{z_j})+\varepsilon \le \oplus _{x\in X}m(f_{n_i}^x)+\varepsilon \le m(f_{k}^{\tilde{x}})+3\varepsilon . \end{aligned}$$

Making $k\rightarrow \infty$, we get

$$\begin{aligned} m(h) \le \tilde{m}(g_{\tilde{x}}^{h(\tilde{x})}) +3\varepsilon =F(h(\tilde{x}),\tilde{x}) +3\varepsilon \le \bigoplus _{x\in X}F(h(x),x) +3\varepsilon . \end{aligned}$$

As $\varepsilon$ is arbitrary, we conclude the proof. $\square$

The next theorem corresponds to Theorem 1 of Ref. [16] in the present setting. It contains the claim in Theorem 1.2.

Theorem 6.11

For each $\lambda \in U(X, \mathbb {R}_{\max })$, consider the functional $m_{\lambda }: C(X,\mathbb {R}) \rightarrow \mathbb {R}_{\max }$ defined by

$$\begin{aligned} m_{\lambda }(h):=\bigoplus _{x \in X} \lambda (x) \odot h(x), \end{aligned}$$

for any $h \in C(X,\mathbb {R})$. Then,

• 1. The map $\gamma :U(X, \mathbb {R}_{\max }) \rightarrow C^{*}(X,\mathbb {R})$ defined by $\gamma (\lambda )= m_{\lambda }$ is a max-plus isomorphism between $U(X, \mathbb {R}_{\max })$ and $C^{*}(X,\mathbb {R})$;

• 2. The function $\lambda \in U(X, \mathbb {R}_{\max })$ is always bounded from above by $m{_\lambda }(0)$ and $m_\lambda \in I(X)$ is equivalent to $\max _{x \in X} \lambda (x)=0$.

Proof

(1) Given $\lambda \in U(X, \mathbb {R}_{\max })$, let us prove that $m_\lambda \in C^{*}(X,\mathbb {R})$. We have

$$\begin{aligned} m_\lambda (a\odot f) = \bigoplus _{x \in X} \lambda (x) \odot a\odot f(x) = a\odot \bigoplus _{x \in X} \lambda (x) \odot f(x) = a\odot m_\lambda ( f) \end{aligned}$$

and

$$\begin{aligned} m_\lambda (f\oplus g)= & {} \bigoplus _{x \in X} \lambda (x) \odot (f(x)\oplus g(x)) \\ {}= & {} \left( \bigoplus _{x \in X} \lambda (x) \odot f(x)\right) \oplus \left( \bigoplus _{x \in X} \lambda (x) \odot g(x)\right) = m_\lambda ( f)\oplus m_\lambda (g). \end{aligned}$$

We claim that $\gamma$ is surjective. From Lemma 6.10, we know that for any $m\in C^{*}(X,\mathbb {R})$ we have

$$\begin{aligned} m(h) = \bigoplus _{x \in X} F(h(x),x )= \bigoplus _{x \in X} F(0,x) \odot h(x). \end{aligned}$$

Defining $\lambda (x):= F(0,x)$, which is u.s.c. from Lemma  6.10, we obtain $m=m_{\lambda }$.

We claim that $\gamma$ is injective. Given $\lambda \in U(X, \mathbb {R}_{\max })$ and $m=m_\lambda$, let F(0, x) as given above. We will prove that $\lambda (x) = F(0,x)$. Indeed, for a fixed $x_0$, consider the standard continuous approximation $(f_n)$ of $g^{0}_{x_0}$. As $\lambda$ is u.s.c., we have

$$\begin{aligned} \lambda (x_0) = \lim _{n\rightarrow \infty } \bigoplus _{x\in X} \lambda (x)\odot f_n(x) \end{aligned}$$

Then

$$\begin{aligned} \lambda (x_0) = \lim _{n\rightarrow \infty } \bigoplus _{x\in X} \lambda (x)\odot f_n(x) = \lim _{n\rightarrow \infty } m_\lambda ( f_n) =\tilde{m}(g^{0}_{x_0}) = F(0,x_0). \end{aligned}$$

We need also to show that $\gamma$ is max-plus linear. To see that take,

$$\begin{aligned} \gamma (a \odot \lambda )(h)= \bigoplus _{x \in X} a \odot \lambda (x) \odot h(x)= a \odot \bigoplus _{x \in X}\lambda (x) \odot h(x)= a \odot \gamma (\lambda )(h), \end{aligned}$$

and

$$\begin{aligned} \gamma (\lambda \oplus \lambda ')(h)= & {} \bigoplus _{x \in X} (\lambda (x) \oplus \lambda '(x)) \odot h(x)=\\= & {} \left( \bigoplus _{x \in X} \lambda (x) \odot h(x) \right) \oplus \left( \bigoplus _{x \in X} \lambda '(x) \odot h(x)\right) =\gamma (\lambda )(h) \oplus \gamma (\lambda ')(f), \end{aligned}$$

for all $h \in C(X, \mathbb {R})$.

(2) We notice that $m_\lambda (0) = \bigoplus _{x\in X} \lambda (x)\odot 0(x) = \bigoplus _{x\in X} \lambda (x)$, thus $\lambda$ is always bounded by $m_\lambda (0)$. Furthermore $\bigoplus _{x \in X} \lambda (x)=0$ if and only if $m_{\lambda }(0) = 0$. $\square$

Remark 6.12

The formula

$$\begin{aligned} m_{\lambda }(h):=\bigoplus _{x \in X} \lambda (x) \odot h(x), \end{aligned}$$

for any $h \in C(X,\mathbb {R})$, in Theorem 6.11 can be slightly improved by recalling that, for each $x \in X$ one can define the Dirac delta measure $\delta _{x}: C(X,\mathbb {R}) \rightarrow \mathbb {R}$ by the formula $\delta _{x}(h):= h(x).$ It is obvious that $\delta _{x} \in C^*(X,\mathbb {R})$ (the density of $\delta _{x}$ is $g^{0}_{x}$). Thus, we can see the previous formula as a kind of Choquet’s theorem (see [29, Theorem Choquet, Pg. 14]), where any idempotent measure is a max-plus combination of Dirac delta measures $\delta _{x}$ with coefficients $\lambda (x)$, that is, $m_{\lambda }(h)=\bigoplus _{x \in X} \lambda (x) \odot \delta _{x}(h),$ for any $f \in C(X,\mathbb {R})$ or,

$$\begin{aligned} m_{\lambda }=\bigoplus _{x \in X} \lambda (x) \odot \delta _{x}. \end{aligned}$$

1.2 Upper semi-continuous envelope

Definition 6.13

Let $\mathcal {B}(X, \mathbb {R}_{\max })$ be the set of bounded functions in the sense of $\rho$, that is, $f: X \rightarrow \mathbb {R}_{\max }$ is bounded if there exists $K>0$ such that $d_{\rho }(f, -\infty )=\max _{x \in X}\rho (f(x), -\infty ) \le K$.

We notice that f is bounded with respect to $\rho$ if, and only if, it is bounded from above in the usual sense because $\rho (f(x), -\infty ) \le K \Leftrightarrow \exp (f(x))\le K \Leftrightarrow f(x) \le \ln (K).$ As X is compact any u.s.c. function (see Definition 6.9) is in $\mathcal {B}(X, \mathbb {R}_{\max })$.

Definition 6.14

Given a function $\lambda \in \mathcal {B}(X, \mathbb {R}_{\max })$, we define the upper semi-continuous envelope of $\lambda$ as

$$\begin{aligned} \lambda ^{u.s.e.}(x)=\inf _{\varphi \in C(X, \mathbb {R}), \; \varphi \ge \lambda } \varphi (x), \; \forall x \in X. \end{aligned}$$

In Ref. [16], the lower semi-continuous envelope is used . The main properties of the upper semi-continuous envelope are given in the next lemma.

Proposition 6.15

By considering Definition 6.14,

• 1. If $\lambda \in \mathcal {B}(X, \mathbb {R}_{\max })$ and $\lambda \ne -\infty$ then $\lambda ^{u.s.e.} \in U(X, \mathbb {R}_{\max })$;

• 2. If $\lambda \in U(X, \mathbb {R}_{\max })$ then $\lambda ^{u.s.e.}=\lambda$.

Proof

(1) First, we notice that $\lambda \ne -\infty$ means that there exists $x_0 \in X$ such that $-\infty <\lambda (x_0)\le \lambda ^{u.s.e.}(x_0)$ thus, $\textrm{supp}(\lambda ^{u.s.e.}) \ne \varnothing$. We claim that $\lambda ^{u.s.e.}$ is u.s.c. Indeed, given $x\in X$ and $\varepsilon >0$, let $(x_n)$ be any sequence such that $x_{n}\rightarrow x$. We fix any function $\varphi _0 \in C(X, \mathbb {R})$ such that $\varphi _0 \ge \lambda$. Then we have

$$\begin{aligned} \lambda ^{u.s.e.}(x_n)=\inf _{\varphi \in C(X, \mathbb {R}), \; \varphi \ge \lambda } \varphi (x_n) \le \varphi _0(x_n) \le \varphi _0(x)+\varepsilon \end{aligned}$$

for n big enough. Thus,

$$\begin{aligned} \limsup _{n\rightarrow \infty }\lambda ^{u.s.e.}(x_n) \le \varphi _0(x)+\varepsilon . \end{aligned}$$

If we take the infimum over functions $\varphi _0$ at the right hand side of this inequality, we get

$$\begin{aligned} \limsup _{n\rightarrow \infty }\lambda ^{u.s.e.}(x_n) \le \lambda ^{u.s.e.}(x)+\varepsilon . \end{aligned}$$

As $\varepsilon$ is arbitrary, we conclude that $\lambda ^{u.s.e.}$ is u.s.c.. As $\textrm{supp}( \lambda ^{u.s.e.}) \ne \varnothing$ we have proved that $\lambda ^{u.s.e.} \in U(X, \mathbb {R}_{\max })$.

(2) As X is compact and $\lambda$ is u.s.c., we know that $\lambda$ attain its maximum value, which we will denote by M. Let $x_0$ be a point of X. As $\lambda$ is u.s.c. at $x_0$, for any real c satisfying $1+M>c> \lambda \left( x_0\right)$, there exists $n\in \mathbb {N}$ such that $\lambda (x)< c$ for all $x \in B_{\frac{1}{n}}(x_0)$. Consider the continuous function $\varphi _{c}$ defined by

$$\begin{aligned} \varphi _{c}(y):= \left\{ \begin{array}{ll} 1+M, &{}\quad d(y,x_0)>1/n \\ c+n(1+M-c)d(y,x_0), &{}\quad d(y,x_0)\le 1/n \end{array} \right. , \end{aligned}$$

(18)

and observe that $\varphi _{c} \ge \lambda$. Thus, $\lambda (x_0) \le \lambda ^{u.s.e.} (x_0) \le \varphi _c(x_0) =c.$ As $c>\lambda (x_0)$ is arbitrary, we conclude that $\lambda ^{u.s.e.}(x_0)=\lambda (x_{0})$. $\square$

Proposition 6.16

For each $\lambda \in \mathcal {B}(X, \mathbb {R}_{\max })$, consider the functional $m_{\lambda }: C(X,\mathbb {R}) \rightarrow \mathbb {R}_{\max }$ defined by

$$\begin{aligned} m_{\lambda }(h):=\bigoplus _{x \in X} \lambda (x) \odot h(x), \end{aligned}$$

for any $h \in C(X,\mathbb {R})$. If $\lambda _{1}, \lambda _{2} \in \mathcal {B}(X, \mathbb {R}_{\max })$ and $m_{\lambda _{1}}= m_{\lambda _{2}},$ then $\lambda _{1}^{u.s.e.}= \lambda _{2}^{u.s.e.}$.

Proof

Since $m_{\lambda _{1}}= m_{\lambda _{2}}$, we have

$$\begin{aligned} \bigoplus _{x \in X} \lambda _{1}(x) \odot h(x) = \bigoplus _{x \in X} \lambda _{2}(x) \odot h(x), \end{aligned}$$

for any $h \in C(X,\mathbb {R})$. Given $\varphi \in C(X, \mathbb {R})$ we have $\varphi \ge \lambda _{1}$ if and only if $\varphi \ge \lambda _{2}$. Indeed, suppose $\varphi \ge \lambda _1$ and take $h=-\varphi$. Then we get

$$\begin{aligned} 0 \ge \bigoplus _{x \in X} \lambda _{1}(x) -\varphi (x) = \bigoplus _{x \in X} \lambda _{2}(x) -\varphi (x), \end{aligned}$$

meaning that $\varphi \ge \lambda _{2}$. The reverse argument is analogous.

Finally, fixed $x_0 \in X$, we have

$$\begin{aligned} \lambda _{1}^{u.s.e.}(x_0)=\inf _{\varphi \in C(X, \mathbb {R}), \; \varphi \ge \lambda _{1}} \varphi (x_0) =\inf _{\varphi \in C(X, \mathbb {R}), \; \varphi \ge \lambda _{2}} \varphi (x_0) = \lambda _{2}^{u.s.e.}(x_0). \end{aligned}$$

$\square$

1.3 Maximal extension of idempotent measures

In this section, we present a brief discussion of how the domain of an idempotent measure can be extended from $C(X,\mathbb {R})$ to $\mathcal {B}(X, \mathbb {R}_{\max })$.

Proposition 6.17

Let $\bar{C}(X,\mathbb {R})$ and $\tilde{m}$ as defined in Lemma 6.8 where $m=m_\lambda$. If $g \in \bar{C}(X,\mathbb {R})$ then

$$\begin{aligned} \tilde{m}(g) = \bigoplus _{x\in X} \lambda (x)\odot g(x). \end{aligned}$$

Proof

Let $f_n$ be a nonincreasing sequence of continuous functions converging pointwise to g. Denote by

$$\begin{aligned} \tilde{m}(g) = \lim _{n\rightarrow \infty } \bigoplus _{x\in X} \lambda (x)\odot f_n(x) \end{aligned}$$

and

$$\begin{aligned} M_\lambda (g) = \bigoplus _{x\in X} \lambda (x)\odot g(x). \end{aligned}$$

We want to prove that $\tilde{m}(g) = M_\lambda (g)$. As $f_n\ge g$ we have

$$\begin{aligned} \tilde{m}(g) = \lim _{n\rightarrow \infty } \bigoplus _{x\in X} \lambda (x)\odot f_n(x) \ge \bigoplus _{x\in X} \lambda (x)\odot g(x)=M_\lambda (g). \end{aligned}$$

To prove the equality suppose initially $M_\lambda (g)\ne -\infty$ and by contradiction suppose there exists an $\varepsilon >0$ and a sequence $(x_n)$ in X such that $\lambda (x_n)\odot f_n(x_n) > M_\lambda (g)+\varepsilon .$ We can suppose there exists $x_0\in X$ such that $x_n\rightarrow x_0$.

Case 1: supposing $\lambda (x_0)\odot g(x_0) \ne -\infty$. As $\lambda (x_0)\odot f_n(x_0)$ converges to $\lambda (x_0)\odot g(x_0)$, there exists k such that $\lambda (x_0)\odot f_k(x_0)< \lambda (x_0)\odot g(x_0)+\varepsilon \le M_{\lambda }(g)+\varepsilon$. As $\lambda +f_k$ is u.s.c., we get a $\delta >0$ such that

$$\begin{aligned} d(x,x_0)<\delta \Rightarrow \lambda (x)\odot f_k(x) < M_\lambda (g)+\varepsilon . \end{aligned}$$

As $f_n$ is nonincreasing, we get

$$\begin{aligned}{}[d(x,x_0)<\delta , n\ge k] \Rightarrow \lambda (x)\odot f_n(x) < M_\lambda (g)+\varepsilon , \end{aligned}$$

which is a contradiction because $d(x_n,x_0)<\delta$ for n large enough.

Case 2: supposing $\lambda (x_0)\odot g(x_0) = -\infty$. Then for any natural N, there exists k such that $\lambda (x_0)\odot f_k(x_0)< -N$. As $\lambda +f_k$ is u.s.c., we get a $\delta >0$ such that

$$\begin{aligned} d(x,x_0)<\delta \Rightarrow \lambda (x)\odot f_k(x) <-N. \end{aligned}$$

As $f_n$ is nonincreasing, we get

$$\begin{aligned}{}[d(x,x_0)<\delta , n\ge k] \Rightarrow \lambda (x)\odot f_n(x) < -N, \end{aligned}$$

which is a contradiction because $d(x_n,x_0)<\delta$ for n large enough.

Now we suppose $M_\lambda (g)= -\infty$, which means $\lambda (x)+g(x) = -\infty ,\;\forall \,x\in X$, and by contradiction, suppose there exists a real number L and a sequence $(x_n)$ in X such that $\lambda (x_n)\odot f_n(x_n) > L.$ We can suppose there exists $x_0\in X$ such that $x_n\rightarrow x_0$. Clearly $\lambda (x_0)\odot g(x_0) = -\infty$ and we get a contradiction arguing as in case 2 above. $\square$

From Theorem 6.11, we have a formula for an idempotent measure $m_{\lambda }$, with density $\lambda$, which can be extended to $h \in \mathcal {B}(X, \mathbb {R}_{\max })$ by setting up

$$\begin{aligned} \hat{m}_{\lambda }(h) = \bigoplus _{x \in X} \lambda (x) \odot h(x), \end{aligned}$$

where the supremum is well defined because h and $\lambda$ are bounded from above. The functional $\hat{m}_{\lambda }$ is called the maximal extension of $m_{\lambda }$ (see Refs. [16, Corollary 1], [2, Theorem 2.2], and [18, Pg. 36], for a minimal version). The precise meaning of this name is explained by Lemma 6.8 and by Remark 6.21.

Definition 6.18

Given $\lambda \in \mathcal {B}(X, \mathbb {R}_{\max })$, we define the idempotent integral for $m_{\lambda } \in C^{*}(X,\mathbb {R})$ by the formula

$$\begin{aligned} \int _{X} h(x)\textrm{d}m_{\lambda }(x):= \hat{m}_{\lambda }(h) = \bigoplus _{x \in X} \lambda (x) \odot h(x), \end{aligned}$$

for all $h \in \mathcal {B}(X, \mathbb {R}_{\max })$. Obviously, if $h \in C(X, \mathbb {R})$ then $\int _{X} h(x) \textrm{d}m_{\lambda }(x)= m_{\lambda }(h)$.

Definition 6.19

Given any $A \subset X$, we define the max-plus indicator function of A as the function

$$\begin{aligned} \chi _{A}(x):= \left\{ \begin{array}{ll} 0, &{}\quad x \in A \\ -\infty , &{}\quad x \not \in A \end{array} \right. . \end{aligned}$$

Obviously, $\chi _{A} \in \mathcal {B}(X, \mathbb {R}_{\max })$, for any $A \subset X$, so we can compute the integral of $\chi _{A}$ with respect to an idempotent measure $m_{\lambda }$ by applying the extension $\hat{m}_{\lambda }$

$$\begin{aligned} \int _{X} \chi _{A} d\mu := \hat{m}_{\lambda }(\chi _{A}) = \bigoplus _{x \in X} \lambda (x) \odot \chi _{A}(x) = \bigoplus _{x \in A} \lambda (x). \end{aligned}$$

We recall that $2^X$ is the set of parts of a set X.

Definition 6.20

Consider $m_{\lambda } \in C^{*}(X,\mathbb {R})$ with extension $\hat{m}_{\lambda }$. The correspondence (we use the same $m_{\lambda }$ as symbol if there is no risk of confusion) $m_{\lambda }: 2^X \rightarrow \mathbb {R}$ defined by

$$\begin{aligned} m_{\lambda }(A):= \int _{X} \chi _{A}(x) \textrm{d}m_{\lambda }(x) = \bigoplus _{x \in A} \lambda (x), \end{aligned}$$

for all $A \in 2^X$ is called the set idempotent measure (see Refs. [18, Corollary 1], [2, Definition 2.1] for cost measures, [1, Section 3] and [11, Definition 1]).

Remark 6.21

By item (3) of Theorem 6.11, if $\lambda \in U(X, \mathbb {R}_{\max })$ and $\lambda _1 \in \mathcal {B}(X, \mathbb {R}_{\max })$ satisfy $m_{\lambda _{1}}= m_{\lambda }$ then $\lambda = \lambda _{1}^{u.s.e.} \ge \lambda _{1}$. Thus,

$$\begin{aligned} m_{\lambda }(A)= \bigoplus _{x \in A} \lambda (x) \ge \bigoplus _{x \in A} \lambda _{1}(x)=m_{\lambda _{1}}(A), \end{aligned}$$

for all $A \in 2^X$. So, the choice of the u.s.c. function $\lambda$ produces an integral bigger or equal to any other choice $\lambda _1$, which explains the meaning of the name “maximal”.

Proposition 6.22

Let $m_{\lambda } \in C^{*}(X,\mathbb {R})$ with extension $\hat{m}_{\lambda }$. Then,

• 1. $m_{\lambda }(\varnothing )=-\infty$;

• 2. $m_{\lambda }(A \cup B)= m_{\lambda }(A ) \oplus m_{\lambda }(B)$ for any $A,B \in 2^X$.

Proof

(1) As usual in the set function theory, we assume that, for any function $f: X \rightarrow \mathbb {R}_{\max }$ the supremum over an empty set is the smallest value in $\mathbb {R}_{\max }$, that is, $-\infty$. Thus, $m_{\lambda }(\varnothing )=\bigoplus _{x \in \varnothing } \lambda (x) = -\infty$.

(2)

$$\begin{aligned} m_{\lambda }(A \cup B)= & {} \int _{X} \chi _{_{A \cup B}}(x) \textrm{d}m_{\lambda }(x)= \hat{m}_{\lambda }(\chi _{_{A \cup B}}) = \bigoplus _{x \in X} \lambda (x) \odot \chi _{_{A \cup B}}(x) \\= & {} \bigoplus _{x \in X} \lambda (x) \odot ( \chi _{_{A}}(x) \oplus \chi _{_{B}}(x))\\{} & {} = \bigoplus _{x \in B} \lambda (x) \oplus \bigoplus _{x \in A} \lambda (x) = m_{\lambda }(A ) \oplus m_{\lambda }(B). \end{aligned}$$

$\square$