[proofplan] We prove the two inequalities separately. The upper bound compares the information in a finite measurable partition with the number of orbit names allowed by a finite open cover, using regularity of Borel probability measures on compact metric spaces. The lower bound constructs invariant measures from maximal separated orbit sets: uniform measures on separated sets are averaged along orbits, weak* limits are invariant, and a small-diameter partition with null boundary converts separated orbit names into partition entropy. Letting the time length and then the spatial scale vary gives equality. [/proofplan] [step:Define the entropy objects used in the proof] Fix a compatible metric $d:X\times X\to[0,\infty)$ on the compact [metrizable space](/page/Metrizable%20Space) $X$. For $n\in\mathbb N$, define the Bowen metric $d_n:X\times X\to[0,\infty)$ by \begin{align*} d_n(x,y):=\max_{0\le k\le n-1}d(T^k x,T^k y). \end{align*} For $\varepsilon>0$, let $s_n(\varepsilon)$ denote the maximal cardinality of an $(n,\varepsilon)$-separated set in $X$, meaning a set $E\subset X$ such that $d_n(x,y)>\varepsilon$ whenever $x,y\in E$ and $x\ne y$. Define \begin{align*} h_{\mathrm{sep}}(T,\varepsilon):=\limsup_{n\to\infty}\frac{1}{n}\log s_n(\varepsilon). \end{align*} By the separated-set characterization of topological entropy, applied to the compatible metric $d$ on the [compact space](/page/Compact%20Space) $X$, \begin{align*} h_{\mathrm{top}}(T)=\lim_{\varepsilon\downarrow 0}h_{\mathrm{sep}}(T,\varepsilon). \end{align*} If $\mathcal U$ is a finite open cover of $X$, define \begin{align*} \mathcal U_0^{n-1}:=\bigvee_{k=0}^{n-1}T^{-k}\mathcal U, \end{align*} where the join of covers consists of all sets $U_0\cap T^{-1}U_1\cap\cdots\cap T^{-(n-1)}U_{n-1}$ with $U_k\in\mathcal U$. Let $N(\mathcal U_0^{n-1})$ denote the least cardinality of a subcover of $\mathcal U_0^{n-1}$. If $\mu\in\mathcal M_T(X)$ and $\mathcal P=\{P_1,\dots,P_m\}$ is a finite Borel partition of $X$, define \begin{align*} H_\mu(\mathcal P):=-\sum_{i=1}^m\mu(P_i)\log\mu(P_i), \end{align*} with the convention $0\log 0=0$. Define the joined partition \begin{align*} \mathcal P_0^{n-1}:=\bigvee_{k=0}^{n-1}T^{-k}\mathcal P. \end{align*} The entropy of $T$ relative to $\mathcal P$ is \begin{align*} h_\mu(T,\mathcal P):=\lim_{n\to\infty}\frac{1}{n}H_\mu(\mathcal P_0^{n-1}), \end{align*} where the limit exists by subadditivity. Finally, \begin{align*} h_\mu(T):=\sup_{\mathcal P}h_\mu(T,\mathcal P), \end{align*} where the supremum is over all finite Borel partitions of $X$. [/step] [step:Bound every measure entropy by topological entropy] Fix $\mu\in\mathcal M_T(X)$ and a finite Borel partition $\mathcal P=\{P_1,\dots,P_m\}$ of $X$. We prove \begin{align*} h_\mu(T,\mathcal P)\le h_{\mathrm{top}}(T). \end{align*} Let $\alpha>0$. By regularity of the Borel probability measure $\mu$ on the compact [metric space](/page/Metric%20Space) $X$, for each atom $P_i$ choose a compact set $K_i\subset P_i$ such that \begin{align*} \mu(P_i\setminus K_i)<\frac{\alpha}{m}. \end{align*} Set $K:=K_1\cup\dots\cup K_m$. Since the finitely many compact sets $K_i$ are pairwise disjoint, choose open sets $U_i\subset X$ with $K_i\subset U_i$ such that $U_i\cap U_l=\varnothing$ whenever $i\ne l$. Define the finite open cover \begin{align*} \mathcal U:=\{U_1,\dots,U_m,X\setminus K\}. \end{align*} This cover is fixed after $\alpha$ and $\mathcal P$ have been chosen. Let $\mathcal R=\{R_0,R_1,\dots,R_m\}$ be the finite Borel partition defined by $R_i=K_i$ for $1\le i\le m$ and $R_0=X\setminus K$. For each $x\in K_i$, membership in $\mathcal R$ determines membership in $\mathcal P$. Thus the only possible disagreement between the one-step information in $\mathcal P$ and $\mathcal R$ occurs on $R_0=X\setminus K$, whose measure satisfies \begin{align*} \mu(R_0)\le\sum_{i=1}^m\mu(P_i\setminus K_i)<\alpha. \end{align*} Let $H_2:[0,1]\to[0,\log 2]$ denote the binary entropy function $H_2(t)=-t\log t-(1-t)\log(1-t)$, with the convention $0\log 0=0$. [claim:Control conditional entropy by an exceptional set] If a finite partition $\mathcal A$ with at most $M$ atoms is determined by a finite partition $\mathcal B$ outside a Borel set $E\subset X$ with $\mu(E)\le\alpha$, then \begin{align*} H_\mu(\mathcal A\mid\mathcal B)\le H_2(\alpha)+\alpha\log M. \end{align*} [/claim] [proof] Let $I_E=\{E,X\setminus E\}$ be the two-atom partition generated by $E$. The chain rule for finite conditional entropy gives \begin{align*} H_\mu(\mathcal A\mid\mathcal B)\le H_\mu(I_E\mid\mathcal B)+H_\mu(\mathcal A\mid\mathcal B\vee I_E). \end{align*} The monotonicity principle that conditioning reduces finite entropy gives $H_\mu(I_E\mid\mathcal B)\le H_\mu(I_E)\le H_2(\alpha)$. On $X\setminus E$, the atom of $\mathcal A$ is determined by the atom of $\mathcal B$, so the remaining conditional entropy is supported on $E$ and is at most $\mu(E)\log M\le\alpha\log M$. Combining these two estimates proves the claim. [/proof] Applying the claim to $\mathcal A=\mathcal P$, $\mathcal B=\mathcal R$, $E=R_0$, and $M=m$ gives \begin{align*} H_\mu(\mathcal P\mid\mathcal R)\le H_2(\alpha)+\alpha\log m. \end{align*} By the chain rule for finite partition entropy and $T$-invariance of $\mu$, \begin{align*} H_\mu(\mathcal P_0^{n-1})\le H_\mu(\mathcal R_0^{n-1})+\sum_{k=0}^{n-1}H_\mu(T^{-k}\mathcal P\mid T^{-k}\mathcal R) \end{align*} and therefore \begin{align*} H_\mu(\mathcal P_0^{n-1})\le H_\mu(\mathcal R_0^{n-1})+n\bigl(H_2(\alpha)+\alpha\log m\bigr). \end{align*} [claim:Code the refined partition by the open-cover name and exceptional times] For every $n\in\mathbb N$, \begin{align*} H_\mu(\mathcal R_0^{n-1})\le \log N(\mathcal U_0^{n-1})+nH_2(\alpha). \end{align*} [/claim] [proof] Let $\mathcal V_n\subset\mathcal U_0^{n-1}$ be a subcover with $|\mathcal V_n|=N(\mathcal U_0^{n-1})$, and choose a measurable tie-breaking map $C_n:X\to\mathcal V_n$ such that $x\in C_n(x)$ for every $x\in X$. For $0\le k\le n-1$, define the Borel set $E_k:=T^{-k}R_0$ and the two-atom partition $I_k:=\{E_k,X\setminus E_k\}$. Since $\mu$ is $T$-invariant, $\mu(E_k)=\mu(R_0)<\alpha$, and hence $H_\mu(I_k)\le H_2(\alpha)$. We claim that the partition $\mathcal R_0^{n-1}$ is determined by the finite data consisting of $C_n$ and the exceptional indicators $I_0,\dots,I_{n-1}$. Indeed, write a selected cover element as \begin{align*} V=V_0\cap T^{-1}V_1\cap\cdots\cap T^{-(n-1)}V_{n-1}, \end{align*} with each $V_k\in\mathcal U$. If $T^k x\in R_0$, the $k$th symbol of the $\mathcal R$-name is $0$. If $T^k x\notin R_0$, then $T^k x\in K_i$ for a unique $i\ge1$. Because the open sets $U_1,\dots,U_m$ are pairwise disjoint and $K_i\subset U_i$, membership of $T^k x$ in $V_k$ determines this unique index $i$. Thus no additional information is needed. Therefore the entropy of $\mathcal R_0^{n-1}$ is bounded by the entropy of $C_n$ plus the entropy of the joined indicator partition $I_0\vee\cdots\vee I_{n-1}$. The [random variable](/page/Random%20Variable) $C_n$ takes at most $N(\mathcal U_0^{n-1})$ values, so its entropy is at most $\log N(\mathcal U_0^{n-1})$. Subadditivity of finite entropy gives \begin{align*} H_\mu(I_0\vee\cdots\vee I_{n-1})\le\sum_{k=0}^{n-1}H_\mu(I_k)\le nH_2(\alpha). \end{align*} Combining the two bounds proves the claim. [/proof] Combining the two estimates and dividing by $n$ gives \begin{align*} \frac{1}{n}H_\mu(\mathcal P_0^{n-1})\le \frac{1}{n}\log N(\mathcal U_0^{n-1})+2H_2(\alpha)+\alpha\log m. \end{align*} Taking $\limsup_{n\to\infty}$, using that $\mathcal U$ is fixed, and then using the open-cover definition of topological entropy as the supremum over finite open covers gives \begin{align*} h_\mu(T,\mathcal P)\le h_{\mathrm{top}}(T)+2H_2(\alpha)+\alpha\log m. \end{align*} Letting $\alpha\downarrow0$ yields \begin{align*} h_\mu(T,\mathcal P)\le h_{\mathrm{top}}(T). \end{align*} Since $\mathcal P$ was arbitrary, \begin{align*} h_\mu(T)\le h_{\mathrm{top}}(T). \end{align*} Taking the supremum over $\mu\in\mathcal M_T(X)$ gives \begin{align*} \sup_{\mu\in\mathcal M_T(X)}h_\mu(T)\le h_{\mathrm{top}}(T). \end{align*} [guided] Fix an invariant Borel probability measure $\mu$ and a finite Borel partition $\mathcal P=\{P_1,\dots,P_m\}$ of $X$. The goal is to compare the measurable orbit names from $\mathcal P$ with the orbit names allowed by one fixed finite open cover. The cover must be fixed before $n$ varies, because topological entropy controls the growth of $N(\mathcal U_0^{n-1})$ for a fixed cover $\mathcal U$. Choose $\alpha>0$. Since $X$ is compact metric, every Borel probability measure on $X$ is regular. Therefore, for each $1\le i\le m$, choose a compact set $K_i\subset P_i$ such that \begin{align*} \mu(P_i\setminus K_i)<\frac{\alpha}{m}. \end{align*} Define the compact set $K\subset X$ by $K=K_1\cup\dots\cup K_m$. The compact sets $K_i$ are pairwise disjoint because the $P_i$ are disjoint. Since $X$ is metric and there are only finitely many of them, choose pairwise disjoint open sets $U_i\subset X$ with $K_i\subset U_i$ for $1\le i\le m$. Define the finite open cover \begin{align*} \mathcal U:=\{U_1,\dots,U_m,X\setminus K\}. \end{align*} This cover is the topological object that will encode the non-exceptional part of the partition names. Define the finite Borel partition $\mathcal R=\{R_0,R_1,\dots,R_m\}$ by $R_i=K_i$ for $1\le i\le m$ and $R_0=X\setminus K$. If $x\in R_i$ with $i\ge1$, then $x\in P_i$, so the $\mathcal R$-atom determines the $\mathcal P$-atom. The only possible failure of determination is the exceptional set $R_0$, and finite subadditivity gives \begin{align*} \mu(R_0)\le\sum_{i=1}^m\mu(P_i\setminus K_i)<\alpha. \end{align*} Let $H_2:[0,1]\to[0,\log 2]$ denote the binary entropy function $H_2(t)=-t\log t-(1-t)\log(1-t)$, with $0\log0=0$. We now prove the conditional-entropy estimate used at one time. Let $\mathcal A$ be a finite partition with at most $M$ atoms, let $\mathcal B$ be a finite partition, and let $E\subset X$ be a Borel set with $\mu(E)\le\alpha$. Assume that $\mathcal A$ is determined by $\mathcal B$ on $X\setminus E$. Define the two-atom partition $I_E=\{E,X\setminus E\}$. The chain rule and monotonicity of conditional entropy give \begin{align*} H_\mu(\mathcal A\mid\mathcal B)\le H_\mu(I_E\mid\mathcal B)+H_\mu(\mathcal A\mid\mathcal B\vee I_E). \end{align*} [Conditioning reduces entropy](/theorems/1652), so \begin{align*} H_\mu(I_E\mid\mathcal B)\le H_\mu(I_E)\le H_2(\alpha). \end{align*} On $X\setminus E$, the atom of $\mathcal A$ is already determined by the atom of $\mathcal B$, so no conditional entropy remains there. On $E$, there are at most $M$ possible atoms of $\mathcal A$, so the contribution is at most $\mu(E)\log M\le\alpha\log M$. Hence \begin{align*} H_\mu(\mathcal A\mid\mathcal B)\le H_2(\alpha)+\alpha\log M. \end{align*} Applying this with $\mathcal A=\mathcal P$, $\mathcal B=\mathcal R$, $E=R_0$, and $M=m$ gives \begin{align*} H_\mu(\mathcal P\mid\mathcal R)\le H_2(\alpha)+\alpha\log m. \end{align*} For $n\in\mathbb N$, the chain rule for finite partition entropy gives \begin{align*} H_\mu(\mathcal P_0^{n-1})\le H_\mu(\mathcal R_0^{n-1})+\sum_{k=0}^{n-1}H_\mu(T^{-k}\mathcal P\mid T^{-k}\mathcal R). \end{align*} Since $\mu$ is $T$-invariant, the entropy of the pulled-back conditional partition is unchanged: \begin{align*} H_\mu(T^{-k}\mathcal P\mid T^{-k}\mathcal R)=H_\mu(\mathcal P\mid\mathcal R). \end{align*} Therefore \begin{align*} H_\mu(\mathcal P_0^{n-1})\le H_\mu(\mathcal R_0^{n-1})+n\bigl(H_2(\alpha)+\alpha\log m\bigr). \end{align*} It remains to bound $H_\mu(\mathcal R_0^{n-1})$ by the open-cover complexity. Let $\mathcal V_n\subset\mathcal U_0^{n-1}$ be a subcover with $|\mathcal V_n|=N(\mathcal U_0^{n-1})$. Since $\mathcal V_n$ is finite, define a measurable tie-breaking map $C_n:X\to\mathcal V_n$ by fixing an ordering $\mathcal V_n=\{V_1,\dots,V_N\}$ and setting $C_n(x)=V_l$ for the least $l$ such that $x\in V_l$. Each fibre of $C_n$ is Borel because it is obtained from finitely many Borel cover elements by intersections and differences. For $0\le k\le n-1$, define $E_k=T^{-k}R_0$ and $I_k=\{E_k,X\setminus E_k\}$. Invariance gives $\mu(E_k)=\mu(R_0)<\alpha$, hence \begin{align*} H_\mu(I_k)\le H_2(\alpha). \end{align*} We claim that the atom of $\mathcal R_0^{n-1}$ containing $x$ is determined by $C_n(x)$ together with the atoms of $I_0,\dots,I_{n-1}$ containing $x$. Write the selected cover element in the form \begin{align*} C_n(x)=V_0\cap T^{-1}V_1\cap\cdots\cap T^{-(n-1)}V_{n-1}, \end{align*} with $V_k\in\mathcal U$. If $x\in E_k$, then $T^k x\in R_0$, so the $k$th $\mathcal R$-symbol is $0$. If $x\notin E_k$, then $T^k x\in K_i$ for a unique $i\ge1$. Because the sets $U_1,\dots,U_m$ are pairwise disjoint and $K_i\subset U_i$, the condition $T^k x\in V_k$ identifies this unique index $i$. Thus the selected cover element plus the exceptional indicators determines every symbol in the $\mathcal R$-name. Consequently, the entropy of $\mathcal R_0^{n-1}$ is bounded by the entropy of the finite-valued map $C_n$ plus the entropy of the joined indicator partition $I_0\vee\cdots\vee I_{n-1}$. The map $C_n$ has at most $N(\mathcal U_0^{n-1})$ values, so \begin{align*} H_\mu(C_n)\le \log N(\mathcal U_0^{n-1}). \end{align*} Subadditivity of finite entropy gives \begin{align*} H_\mu(I_0\vee\cdots\vee I_{n-1})\le\sum_{k=0}^{n-1}H_\mu(I_k)\le nH_2(\alpha). \end{align*} Hence \begin{align*} H_\mu(\mathcal R_0^{n-1})\le \log N(\mathcal U_0^{n-1})+nH_2(\alpha). \end{align*} Combining this estimate with the previous estimate for $H_\mu(\mathcal P_0^{n-1})$ and dividing by $n$ gives \begin{align*} \frac{1}{n}H_\mu(\mathcal P_0^{n-1})\le \frac{1}{n}\log N(\mathcal U_0^{n-1})+2H_2(\alpha)+\alpha\log m. \end{align*} Because $\mathcal U$ is fixed, the open-cover definition of topological entropy implies \begin{align*} h_\mu(T,\mathcal P)\le h_{\mathrm{top}}(T)+2H_2(\alpha)+\alpha\log m. \end{align*} Finally let $\alpha\downarrow0$. Since $H_2(\alpha)\to0$, we obtain \begin{align*} h_\mu(T,\mathcal P)\le h_{\mathrm{top}}(T). \end{align*} Taking the supremum over all finite Borel partitions gives $h_\mu(T)\le h_{\mathrm{top}}(T)$, and then taking the supremum over $\mu\in\mathcal M_T(X)$ gives the desired upper bound. [/guided] [/step] [step:Build invariant measures from separated orbit sets] Fix $\varepsilon>0$ and fix a real number $L$ satisfying $0\le L<h_{\mathrm{sep}}(T,\varepsilon)$, with the convention that such $L$ is arbitrary when $h_{\mathrm{sep}}(T,\varepsilon)=\infty$. By the definition of the limit superior, choose integers $n_j\to\infty$ such that \begin{align*} \frac{1}{n_j}\log s_{n_j}(\varepsilon)\ge L \end{align*} for every $j$. For each $j$, choose an $(n_j,\varepsilon)$-separated set $E_j\subset X$ with \begin{align*} |E_j|=s_{n_j}(\varepsilon). \end{align*} Define the uniform probability measure on $E_j$ by \begin{align*} \sigma_j:=\frac{1}{|E_j|}\sum_{x\in E_j}\delta_x. \end{align*} For a Borel probability measure $\rho$ on $X$, define the pushforward measure $T_*\rho$ on the Borel subsets of $X$ by $(T_*\rho)(A)=\rho(T^{-1}A)$. Define the orbit-averaged probability measure \begin{align*} \nu_j:=\frac{1}{n_j}\sum_{k=0}^{n_j-1}T_*^k\sigma_j. \end{align*} By weak star compactness of probability measures, the space of Borel probability measures on the compact metric space $X$ is weak* compact. Passing to a subsequence, still indexed by $j$, there exists a Borel probability measure $\mu$ on $X$ such that $\nu_j$ converges weak star to $\mu$. We verify that $\mu$ is $T$-invariant. Let \begin{align*} f:X\to\mathbb R \end{align*} be continuous. Then \begin{align*} \int_X f\,d(T_*\nu_j)-\int_X f\,d\nu_j=\frac{1}{n_j}\left(\int_X f\,d(T_*^{n_j}\sigma_j)-\int_X f\,d\sigma_j\right). \end{align*} Taking absolute values and using $| \int_X f\,d\rho |\le \|f\|_\infty$ for every probability measure $\rho$ gives \begin{align*} \left|\int_X f\,d(T_*\nu_j)-\int_X f\,d\nu_j\right|\le \frac{2\|f\|_\infty}{n_j}. \end{align*} The right-hand side tends to $0$. Passing to the weak* limit gives \begin{align*} \int_X f\,d(T_*\mu)=\int_X f\,d\mu. \end{align*} Since this holds for every continuous $f:X\to\mathbb R$, we have $T_*\mu=\mu$, so $\mu$ is an invariant measure and $\mu\in\mathcal M_T(X)$. [/step] [step:Choose a small partition whose boundary is invisible to the limit measure] For a Borel set $A\subset X$, let $\partial A:=\overline{A}\setminus A^\circ$ denote its topological boundary in $X$. We choose a finite Borel partition $\mathcal P=\{P_1,\dots,P_m\}$ of $X$ such that every atom has diameter less than $\varepsilon$ and \begin{align*} \mu(\partial P_i)=0 \end{align*} for every $1\le i\le m$. To construct it, cover $X$ by finitely many open balls of radius less than $\varepsilon/3$. Fix one centre $x_a$. The spheres $\partial B(x_a,r)$ for distinct radii $r>0$ are pairwise disjoint. Since $\mu(X)=1$, for each integer $N\ge1$ there can be only finitely many radii $r>0$ with $\mu(\partial B(x_a,r))\ge 1/N$; otherwise finitely many disjoint such spheres would have total measure exceeding $1$. Hence the set of radii with $\mu(\partial B(x_a,r))>0$ is countable. For each chosen centre, select a radius still less than $\varepsilon/3$ and outside this countable exceptional set. Taking the corresponding balls and then disjointizing them by subtracting the preceding balls gives a finite Borel partition. Boundaries of the resulting atoms are contained in finite unions of spheres of $\mu$-measure zero, and the diameters remain less than $\varepsilon$. [/step] [step:Convert separated orbit names into partition entropy] For each $j$, the map \begin{align*} \pi_j:E_j\to \mathcal P_0^{n_j-1},\qquad x\mapsto \text{the atom of }\mathcal P_0^{n_j-1}\text{ containing }x \end{align*} is injective. Indeed, if $x,y\in E_j$ have the same $\mathcal P_0^{n_j-1}$-name, then for each $0\le k\le n_j-1$ the points $T^k x$ and $T^k y$ lie in the same atom of $\mathcal P$, whose diameter is less than $\varepsilon$. Therefore \begin{align*} d_{n_j}(x,y)<\varepsilon. \end{align*} Since $E_j$ is $(n_j,\varepsilon)$-separated, this forces $x=y$. Thus the partition $\mathcal P_0^{n_j-1}$ separates all atoms of the uniform measure $\sigma_j$, and hence \begin{align*} H_{\sigma_j}(\mathcal P_0^{n_j-1})=\log |E_j|=\log s_{n_j}(\varepsilon). \end{align*} Fix $q\in\mathbb N$. We prove the block estimate used to average orbit names. For each offset $r\in\{0,\dots,q-1\}$, let \begin{align*} B_{j,r}:=\{t\in\{0,\dots,n_j-q\}:t\equiv r\pmod q\}. \end{align*} The intervals $\{t,t+1,\dots,t+q-1\}$ with $t\in B_{j,r}$ are disjoint. They cover all times in $\{0,\dots,n_j-1\}$ except an initial remainder of length at most $q$ and a terminal remainder of length less than $q$. The joined partition over each remainder has entropy at most its length times $\log m$, so [subadditivity of entropy](/theorems/1634) under refinement gives \begin{align*} H_{\sigma_j}(\mathcal P_0^{n_j-1})\le \sum_{t\in B_{j,r}}H_{\sigma_j}(T^{-t}\mathcal P_0^{q-1})+2q\log m. \end{align*} Averaging this inequality over $r=0,\dots,q-1$ and using the identity $H_{\sigma_j}(T^{-t}\mathcal P_0^{q-1})=H_{T_*^t\sigma_j}(\mathcal P_0^{q-1})$ yields \begin{align*} H_{\sigma_j}(\mathcal P_0^{n_j-1})\le \frac{1}{q}\sum_{t=0}^{n_j-1}H_{T_*^t\sigma_j}(\mathcal P_0^{q-1})+2q\log m. \end{align*} The concavity of finite partition entropy as a function of the underlying probability measure, applied to the average measure $\nu_j=n_j^{-1}\sum_{t=0}^{n_j-1}T_*^t\sigma_j$, gives \begin{align*} \frac{1}{n_j}\sum_{t=0}^{n_j-1}H_{T_*^t\sigma_j}(\mathcal P_0^{q-1})\le H_{\nu_j}(\mathcal P_0^{q-1}). \end{align*} Combining these estimates with $H_{\sigma_j}(\mathcal P_0^{n_j-1})=\log s_{n_j}(\varepsilon)$ and dividing by $n_j$ yields \begin{align*} \frac{1}{n_j}\log s_{n_j}(\varepsilon)\le \frac{1}{q}H_{\nu_j}(\mathcal P_0^{q-1})+\frac{2q\log m}{n_j}. \end{align*} [/step] [step:Pass the entropy estimate to the weak star limit] Because $\mu(\partial P_i)=0$ for every atom $P_i$ of $\mathcal P$, every atom of the joined partition $\mathcal P_0^{q-1}$ has $\mu$-null boundary. Indeed, if $A=P_{i_0}\cap T^{-1}P_{i_1}\cap\cdots\cap T^{-(q-1)}P_{i_{q-1}}$ is such an atom, we use the standard topological fact that for a continuous map $S:X\to X$ and a Borel set $B\subset X$ one has $\partial(S^{-1}B)\subset S^{-1}(\partial B)$. Applying this to $S=T^k$ and $B=P_{i_k}$ gives \begin{align*} \partial A\subset \bigcup_{k=0}^{q-1}T^{-k}(\partial P_{i_k}). \end{align*} Since $\mu$ is $T$-invariant, $\mu(T^{-k}(\partial P_{i_k}))=\mu(\partial P_{i_k})=0$ for each $k$, and hence $\mu(\partial A)=0$ by finite subadditivity. By the [Portmanteau Theorem](/theorems/1171) for continuity sets, weak* convergence $\nu_j\to\mu$ therefore implies convergence of the atom masses of $\mathcal P_0^{q-1}$. Since the function \begin{align*} \psi:[0,1]\to[0,\infty),\qquad \psi(t):=-t\log t \end{align*} is continuous with $\psi(0)=0$, we obtain \begin{align*} H_{\nu_j}(\mathcal P_0^{q-1})\to H_\mu(\mathcal P_0^{q-1}). \end{align*} Taking the limit superior in the previous estimate and using the defining lower bound for the sequence $n_j$ gives \begin{align*} L\le \frac{1}{q}H_\mu(\mathcal P_0^{q-1}). \end{align*} Now let $q\to\infty$. By the definition of partition entropy, \begin{align*} L\le h_\mu(T,\mathcal P)\le h_\mu(T). \end{align*} Since $\mu\in\mathcal M_T(X)$, this proves \begin{align*} L\le \sup_{\rho\in\mathcal M_T(X)}h_\rho(T). \end{align*} Because $L<h_{\mathrm{sep}}(T,\varepsilon)$ was arbitrary, including arbitrarily large finite $L$ when $h_{\mathrm{sep}}(T,\varepsilon)=\infty$, we obtain \begin{align*} h_{\mathrm{sep}}(T,\varepsilon)\le \sup_{\rho\in\mathcal M_T(X)}h_\rho(T). \end{align*} Finally let $\varepsilon\downarrow0$. Using the separated-set formula for topological entropy, \begin{align*} h_{\mathrm{top}}(T)=\lim_{\varepsilon\downarrow0}h_{\mathrm{sep}}(T,\varepsilon), \end{align*} we conclude \begin{align*} h_{\mathrm{top}}(T)\le \sup_{\rho\in\mathcal M_T(X)}h_\rho(T). \end{align*} [/step] [step:Combine the two inequalities] The upper bound proved \begin{align*} \sup_{\mu\in\mathcal M_T(X)}h_\mu(T)\le h_{\mathrm{top}}(T). \end{align*} The lower bound proved \begin{align*} h_{\mathrm{top}}(T)\le \sup_{\mu\in\mathcal M_T(X)}h_\mu(T). \end{align*} Therefore \begin{align*} h_{\mathrm{top}}(T)=\sup_{\mu\in\mathcal M_T(X)}h_\mu(T). \end{align*} This is the variational principle for topological entropy. [/step]