Projective limit of spaces of probability measures is bijective to the space of probability measures on a projective limit.

Solution 1:

Considering it has been 3 months, I will post the proof I came up with, appealing to the work of Choksi [1]. I will present the whole proof, and not just the one showing bijectivity, because I have already typed it up in TEX as such.

We first aim to prove that $\Phi:M_\varphi^1(K)\to \varprojlim_jM^1_{\varphi_j}(K_j)$ given by $\mu\mapsto (\pi_{j*}\mu)_{j\in I}$ is a bijective mapping between compact and convex sets. We work under assumption that $\pi_{ij}$ is surjective for each $i\leq j$. We proceed by first proving the following lemma:

Lemma 1: Let $(K_j,\pi_{ij},I)$ be an inverse system of compact and Hausdorff topological spaces. The projective limit of this system $K:=\varprojlim_jK_j$ is compact.

Proof: We first note that $\prod_{j\in I}K_j$ is compact by Tychonoff's theorem. It thus suffices to prove that $K$ is closed in $\prod_{j\in I}K_j$. To that end fix some $(x_j)_I\in \prod_{j\in I}K_j\backslash K$. There then exists some $i\leq j$ such that $\pi_{ij}x_j\neq x_i$. Fix an open neighbourhood $U\subset K_i$ of $x_i$ and an open neighbourhood $V\subset K_i$ of $\pi_{ij}x_j$ disjoint from $U$. Set $V'=\pi_{ij}^{-1}(V)\subset K_j$. $V'$ is obviously an open neighbourhood of $x_j$. Now define $W=\prod_{k\in I}W_k$ where $W_k=K_k$ for $k\neq i,j$ and $W_j=V'$ and $W_i=U$. By construction $W\cap K=\emptyset$ and $W$ is the product of open sets so is open. Thus $(x_j)_I$ is an interior point, but it is arbitrary, so we can conclude that $K$ is indeed closed and thus compact. Q.E.D

We know that for any compact topological system that the set of invariant probability measures on that system is compact, so we can conclude that each $M^1_{\varphi_j}(K_j)$ is compact. This in conjunction with Lemma 1 means that $M^1_\varphi(K)$ is compact. The same lemma also thus implies that $\varprojlim_j M^1_{\varphi_j}(K_j)$ is compact. We also want $K$ to be none empty, to which end we prove:

Lemma 2: For any $i\in I$ and $x_i\in K_i$ there exists a $x\in K$ such that $\pi_i(x)=x_i$.

Proof: Define for any $j\geq i$, $$L_j=\left\{(\tilde x)_{\ell\in I}\in\prod_{\ell \in I}K_\ell:\pi_i(\tilde x)=x_i,~\pi_{kj}\tilde x_j=\tilde x_k~\text{for all}~k\leq j\right\}.$$ By mirroring the argument showing that $K$ is closed in the proof of lemma 1 it follows that each $L_j$ is closed in the product space, so indeed each $L_j$ is compact as $\prod_{j\in I}K_j$ is compact Hausdorff. Furthermore, by construction, for each $(\tilde x)\in \bigcap_{j\in I}L_j$ we have that $\pi_i \tilde x=x_i$, and for every $k\leq j$ we have $\pi_{kj}\tilde x_j=\tilde x_k$. We thus just need to show that $\bigcap_{j\in I}L_j\neq \emptyset$. We do this by utilizing the fact that in a compact space if a collection of closed sets satisfies the finite intersection property then the intersection of all the sets in the collection is nonempty. Now for any finite $\{j_1,\dots j_n\}\subset I$ we have by the definition of a directed set that there exists some $j\in I$ such that $j_k\leq j$ for all $k\in \{1,\dots,n\}$. Now $\emptyset\neq L_{j}\subset \bigcap_{k=1}^nL_{j_k}$, so we can conclude that every $\{L_j\}_{j\in L}$ is a collection of closed sets with the finite intersection property, so by the above $\bigcap_{j\in I}L_j\neq \emptyset$. This allows us to conclude that $\pi_i(\tilde x)=x_i$ for any $\tilde x\in \bigcap_{j\in I}L_j\subset K$. Q.E.D

The above Lemma actually gives us more than $K$ being non-empty, but actually says that each projection restricted to $K$ is surjective. We now present a simple lemma, but without proof as it is fairly simple.

Lemma 3: The projective limit, $X$, of an inverse system $(X_i,T_{ij},I)$ of linear spaces is a convex subset of the product $X=\prod_{j\in I}X_j$.

Lemma 3 allows us to conclude that each $\varprojlim_jM_{\varphi_j}^1(K_j)$ is indeed convex, so $\Phi$ is a map between non-empty compact, convex spaces.

We must now show that $\Phi$ is bijective, affine and continuous. To do this we appeal to the work of Choksi [1]. Lemma 2 guarantees that our inverse system $(K_i,\pi_{ij},I)$ satisfies the assumptions that Choksi works under. We can associate any $(\mu_i)\in \varprojlim_iM_{\varphi_i}^1(K_i)$ with a inverse system of Baire probability measure spaces $(K_i,\mathcal{B}(K_i),\mu_i,\pi_{ij},I)$. Theorem 2.2 and Theorem 2.3 in [1] guarantee that the inverse limit measure space, $(K,\otimes\mathcal{B}(K_i),\mu)$, of this system exists, and that this is indeed a Baire probability space. That the limit space of probability spaces is again a probability space is not explicitly stated, but this follows simply from the construction where a finitely additive set function on a generating ring of $\otimes\mathcal{B}(K_i)$ is defined in terms of the measures on each $K_i$. Furthermore this same construction gives that $\pi_{i*}\mu=\mu_i$ for all $i\in I$. So we have $\mu\in M^1(K)$ satisfying $\pi_{i*}\mu=\mu_i$, and all that is left is to show that $\mu$ is $\varphi$ invariant. This follows because on any element of the generating semi-ring of cylinder sets, of the form $\cap_{i=1}^n\pi_{j_i}^{-1}(A_{j_i})$ for $A_{j_i}\in \mathcal{B}(K_{j_i})$, we have \begin{align*} \mu\left(\varphi^{-1}\bigcap_{i=1}^n\pi_{j_i}^{-1}(A_{j_i})\right)&=\mu\left(\bigcap_{i=1}^n\varphi^{-1}\pi_{j_i}^{-1}(A_{j_i})\right)\\ &=\mu\left(\bigcap_{i=1}^n\pi_{j_i}^{-1}(\varphi_{j_i}^{-1}A_{j_i})\right)\\ &=\prod_{i=1}^n\mu_{j_i}(\varphi_{j_i}^{-1}(A_{j_i}))) \\ & =\prod_{i=1}^n\mu_{j_i}(A_{j_i})\\ &= \mu\left(\bigcap_{i=1}^n\pi_{j_i}^{-1}(A_{j_i})\right), \end{align*} meaning that $\mu$ is $\varphi$ invariant on a generating semi-ring, so indeed $\mu\in M^1_\varphi(K)$. Thus $\Phi$ is surjective.

To prove that $\Phi$ is injective we suppose $\Phi(\mu)=(\mu_i)_{_i\in I}=\Phi(\nu)$. In particular this means $\pi_i^*\mu=\mu_i=\pi_i^*\nu$, which means that both $\pi$ and $\nu$ agree on the generating semi ring of $\mathcal{B}(K)$, namely cylinder sets of the form $\cap_{i=1}^n\pi_{j_i}^{-1}(A_{j_i})$ for $A_{j_i}\in \mathcal{B}(K_{j_i})$. However the Caratheodory Extension Theorem means that the extension of a finite measure on a generating semi-ring to the $\sigma$ algebra is unique, so we must have $\mu=\nu$.

We now show that $\Phi$ is continuous with continuous inverse. To do so we know that $\Phi$ can easily be thought of as the restriction of a bijective mapping $\Psi:M(K)\to \varprojlim_iM(K_i)$ (The bijectivity of $\Psi$ follows almost identically from the arguments given above.) We claim that $\Psi$ is linear. To see this fix some $\mu,\nu\in M(K)$. Then, for any $i\in I$ and $A\in \mathcal{B}(K_i)$, \begin{align*} (\Psi(\mu+\nu))_i(A)&=\pi_{i*}(\nu+\mu)(A)=\mu(\pi_i^{-1}(A))+\nu(\pi_i^{-1}(A))\\ &=\pi_{i*}\mu(A)+\pi_{i*}\nu(A)=(\Psi(\mu))_i(A)+(\Psi(\nu))_i(A). \end{align*} Similarly, for any $\lambda\in \mathbb C$ we have $$(\Psi\lambda\mu)_i(A)=\pi_*\lambda\mu(A)=\lambda\mu(\pi^{-1}_i(A))=\lambda(\Psi\mu)_i.$$ Thus $\Psi$ is bounded, and so $\Phi$ as its restriction to a map between convex spaces is affine.

We now prove that $\Psi$ is bounded. Define $J_i:C(K_i)\to C(K)$ as the Koopman operator of the projection $\pi_i:K\to K_i$, and $J_{ji}:C(K_i)\to C(K_j)$ as the Koopman operator of $\pi_{ij}$. We can define a mapping $$F:\mathscr L(C(K);C(K))\to X,$$ by $T\mapsto(TJ_i)i$, where $$X=\{(T_i)_{i\in I}:\sup_{i\in I}\|T_i\|<\infty,~\text{and}~T_jJ_{ji}=T_i~\forall i\leq j\}.$$ It can be shown that $\|F(T)\|=\sup_{i\in I}\|TJ_i\|$, which is how we will show that $\Psi$ is bounded. We note that the adjoint of $(J_i)_i\in X$ is in fact $\Psi:M(K)\to \varprojlim_iM(K_i)$, but $(J_i)_i=F(\operatorname{id}),$ and because $\|J_i\|=1$ for all $i\in I$ we get that $\|F(\operatorname{id})\|=1$. Thus $\|\Psi\|=1$, and $\Psi$ is indeed a bounded operator. As the restriction of a bounded operator $\Phi$ is thus continuous. We have also proven that it is a bijective mapping between compact Hausdorff spaces, so the inverse of $\Phi$ is thus also continuous so $\Phi$ is an affine homeomorphism as claimed.

[1] Choksi, J. R., Inverse limits of measure spaces, Proc. Lond. Math. Soc., III. Ser. 8, 321-342 (1958). ZBL0085.04003.

Solution 2:

I know this is an old post, but here's what I came up with while looking at that book. Note: I used different notation than the book on this one. I used $T_i$ to refer to the continuous map $T_i : X_i \to X_i$ instead of $\varphi_i : K_i \to K_i$, and I used $\phi_{i, j} : X_j \to X_i$ to refer to the transition maps.

I do use freely the claim that $(\pi_i)_{i \in I}$ separates points. This might be valuable to check for yourself.

Let $$\left( (X_i, T_i)_{i \in I} , (\phi_{i, j})_{i, j \in I, i \leq j} \right)$$ be an inverse system of compact topological dynamical systems. This inverse system induces a dual inverse system $$\left( \left(\mathcal{M}_{T_i}^1(X_i) \right)_{i \in I} , (\phi_{i, j *})_{i, j \in I, i \leq j} \right)$$ in the category of Choquet simplexes, where $\phi_{i, j *} : \mathcal{M}_{T_j}^1(X_j) \to \mathcal{M}_{T_i}^1(X_i)$ is the continuous affine map \begin{align*} \int_{X_i} f \mathrm{d} \left( \phi_{i, j *} \mu \right) & = \int_{X_j} (f \circ \phi_{i, j}) \mathrm{d} \mu & \left( f \in C(X_i) , \mu \in \mathcal{M}_{T_j}^1(X_j) \right) . \end{align*} Let $(X, T) = \varprojlim (X_i, T_i) $. Then $\mathcal{M}_T^1(X)$ is affinely homeomorphic to $\varprojlim \mathcal{M}_{T_i}^1(X_i)$.

Let $\Phi : \mathcal{M}_T^1(X) \to \varprojlim \mathcal{M}_{T_i}^1(X_i)$ be the map $\Phi(\mu) = (\pi_{i *} \mu)_{i \in I}$. The map is clearly continuous and affine, so it remains to show that $\Phi$ is a bijection. Let $\pi_i : (X, T) \to (X_i, T_i)$ be the factor maps.

Before we prove the map $\Phi$ is bijective, we need to establish a fact about a particular subset of $C(X)$ which we'll be making liberal use of. Consider the set $A = \bigcup_{i \in I} \left\{ f \circ \pi_i : f \in C(X_i) \right\} \subseteq C(X)$. We claim that $A$ is a dense unital $*$-subalgebra of $C(X)$. Clearly $A$ is closed under scalar multiplication and complex conjugation, and contains $1$, so it remains to show that $A$ is closed under addition. Let $f_i \in C(X_i), f_j \in C(X_j)$ for $i, j \in I$. Then $f_i \circ \pi_i = (f_i \circ \phi_{i, k}) \circ \pi_k , f_j \circ \pi_j = (f_j \circ \phi_{j, k}) \circ \pi_k$, where $f_i \circ \phi_{i, k} , f_j \circ \phi_{j, k} \in C(X_k)$, so we can write \begin{align*} (f_i \circ \pi_i) + (f_j \circ \pi_j) & = \left( (f_i \circ \phi_{i, k}) + (f_j \circ \phi_{j, k}) \right) \circ \pi_k & \in \pi_{k *} C(X_k) & \subseteq A , \\ (f_i \circ \pi_i) \cdot (f_j \circ \pi_j) & = \left( (f_i \circ \phi_{i, k}) \cdot (f_j \circ \phi_{j, k}) \right) \circ \pi_k & \in \pi_{k *} C(X_k) & \subseteq A . \end{align*} Therefore, $A$ is a $*$-algebra. Finally, we claim that $A$ is dense in $C(X)$. To see this, we just need to confirm that $A$ separates points. Let $x, y \in X, x \neq y$. Then there exists $i \in I$ such that $\pi_i(x) \neq \pi_i(y)$. Let $f \in C(X_i)$ such that $f(\pi_i(x)) \neq f(\pi_i(y))$. Then $A$ separates points, so we can apply Stone-Weierstrass to confirm that $A$ is a dense $*$-subalgebra of $C(X)$, and in particular that $A$ is a dense subspace of $C(X)$.

Now we can show that the map $\Phi$ is injective. Let $\mu, \nu$ be two $T$-invariant Borel probability measures on $X$ such that $\Phi(\mu) = \Phi(\nu)$, i.e. $\pi_{i *} \mu = \pi_{i *} \nu$ for all $i \in I$. Let $f \in C(X_i)$. Then $$\int_X (f \circ \pi_i) \mathrm{d} \mu = \int_{X_i} f \mathrm{d} (\pi_{i *} \mu) = \int_{X_i} f \mathrm{d} (\pi_{i *} \nu) = \int_X (f \circ \pi_i) \mathrm{d} \nu .$$ Since we've proven that $\int_X g \mathrm{d} \mu = \int_X g \mathrm{d} \nu$ for all $g \in A$, we can appeal to density to conclude that $\int_X g \mathrm{d} \mu = \int_X g \mathrm{d} \nu$ for all $g \in C(X)$, meaning that $\mu = \nu$, telling us that the map $\Phi$ is injective.

Now we prove that the map $\Phi$ is surjective. Let $(\mu_{i})_{i \in I} \in \prod_{i \in I} \mathcal{M}_{T_i}^1(X_i)$ be such that $\phi_{i, j *} \mu_j = \mu_i$ for all $i, j \in I, i \leq j$. Define a measure $\mu$ on $X$ by \begin{align*} \int_X f \circ \pi_i \mathrm{d} \mu & = \int_{X_i} f \mathrm{d} \mu_i & (\forall i \in I , \forall f \in C(X_i)) . \end{align*} Before proceeding, we'll go ahead and check that this $\mu$ is well-defined on $A$. Let $i, j \in I$, and suppose we have $f_i \in C(X_i) , f_j \in C(X_j)$ such that $f_i \circ \pi_i = f_j \circ \pi_j$. Choose $k \in I$ such that $i \leq k , j \leq k$, which exists because $I$ is directed. Then $f_i \circ \phi_{i, k} = f_j \circ \phi_{j, k} \in C(K)$, and \begin{align*} (f_i \circ \phi_{i, k}) \circ \pi_k & = f_i \circ \pi_i , \\ (f_j \circ \phi_{j, k}) \circ \pi_k & = f_j \circ \pi_j . \end{align*} Therefore \begin{align*} \int_{X_k} (f_i \circ \phi_{i, k}) \mathrm{d} \mu_k & = \int_{X_i} f_i \mathrm{d} \left( \phi_{i, k *} \mu_k \right) \\ & = \int_{X_i} f_i \mathrm{d} \mu_i , \end{align*} and a similar computation tells us that $\int_{X_k} (f_i \circ \phi_{i, k}) \mathrm{d} \mu_k = \int_{X_k} (f_j \circ \phi_{j, k}) \mathrm{d} \mu_k = \int_{X_j} f_j \mathrm{d} \mu_j$. Therefore, our definition of $\mu$ above is unambiguous. Since this $\mu$ is a unital positive functional on $A$, we can extend it using density to all of $C(X)$. Therefore we've constructed a Borel probability measure $\mu$ on $X$ such that $\Phi(\mu) = (\mu_i)_{i \in I}$. Finally, to confirm that this $\mu$ is $T$-invariant, it suffices to observe that if $f \in C(X_i)$, then \begin{align*} \int_X (f \circ \pi_i) \circ T \mathrm{d} \mu & = \int_X (f \circ T_i) \circ \pi_i \mathrm{d} \mu \\ & = \int_{X_i} (f \circ T_i) \mathrm{d} \mu_i \\ & = \int_{X_i} f \mathrm{d} \mu_i \\ & = \int_X (f \circ \pi_i) \mathrm{d} \mu . \end{align*} Therefore, we have shown that $\int_X g \circ T \mathrm{d} \mu = \int_X g \mathrm{d} \mu$ for all $g \in A$, and since $A$ is dense in $C(X)$, we can extend this equality to all of $C(X)$.