[proofplan] We exhibit the K-sigma-algebra explicitly: take the sigma-algebra generated by the nonnegative coordinates. Its shifted preimages form the decreasing future-tail filtration, while its forward images recover every coordinate and hence the whole product sigma-algebra. The only substantive point is that the future tail is trivial; this follows from independence of disjoint coordinate blocks and the standard zero-one argument for independent tails. Finally, we check that the K-property is invariant under measure-theoretic isomorphism. [/proofplan]

[step:Choose the future coordinate sigma-algebra] Let $(A,\mathcal{A},\nu)$ be the one-coordinate probability space of the Bernoulli system. We first prove the result for the two-sided Bernoulli shift. Define the product space by \begin{align*} X := A^{\mathbb{Z}}. \end{align*} Define the product sigma-algebra by \begin{align*} \mathcal{B} := \mathcal{A}^{\otimes \mathbb{Z}}. \end{align*} Define the product probability measure by \begin{align*} \mu := \nu^{\otimes \mathbb{Z}}. \end{align*} Define the shift map $\sigma: X \to X$ by $\sigma((x_j)_{j \in \mathbb{Z}}) := (x_{j+1})_{j \in \mathbb{Z}}$. Thus $(X,\mathcal{B},\mu,\sigma)$ is the two-sided Bernoulli shift model, and $\mathcal{B}$ is the product sigma-algebra. All sigma-algebras are considered modulo $\mu$-null sets when verifying the K-property. Let $\mathbb{N} := \{1,2,3,\dots\}$ and let $\mathbb{N}_0 := \{0,1,2,\dots\}$. For each $k \in \mathbb{Z}$, define the coordinate map $X_k: X \to A$ by $X_k(x) = x_k$ for $x \in X$. Let $\mathcal{F} \subset \mathcal{B}$ be the sub-sigma-algebra \begin{align*} \mathcal{F} := \sigma(X_k : k \ge 0). \end{align*} Since $(\sigma x)_k = x_{k+1}$, for every $k \ge 0$ one has $X_k \circ \sigma = X_{k+1}$. Hence \begin{align*} \sigma^{-1}\mathcal{F} = \sigma(X_k \circ \sigma : k \ge 0) = \sigma(X_k : k \ge 1) \subset \sigma(X_k : k \ge 0) = \mathcal{F}. \end{align*} Thus $\mathcal{F}$ satisfies the monotonicity condition required in the definition of the K-property. [/step]

[step:Show that forward translates generate the full product sigma-algebra]For $m \in \mathbb{N}_0$, the sigma-algebra $\sigma^m\mathcal{F}$ is generated by the coordinates $X_k$ with $k \ge -m$. Therefore \begin{align*} \bigvee_{m=0}^{\infty} \sigma^m\mathcal{F} = \sigma(X_k : k \in \mathbb{Z}) = \mathcal{B}. \end{align*} The first equality holds because every coordinate $X_j$ is included once $m \ge -j$ if $j < 0$, while the nonnegative coordinates are already included for $m=0$. The second equality is the definition of the product sigma-algebra on $A^{\mathbb{Z}}$.[/step]

[guided]We recall the objects used in this step so that the guided argument is self-contained. The space is $X=A^{\mathbb{Z}}$, the sigma-algebra is $\mathcal{B}=\mathcal{A}^{\otimes\mathbb{Z}}$, the shift map $\sigma:X\to X$ is given by $\sigma((x_j)_{j\in\mathbb{Z}})=(x_{j+1})_{j\in\mathbb{Z}}$, and the coordinate maps are $X_k:X\to A$, $X_k(x)=x_k$. The future coordinate sigma-algebra is \begin{align*} \mathcal{F}:=\sigma(X_k:k\geq 0). \end{align*} The purpose of this step is to verify that $\mathcal{F}$ is large enough, after applying forward shifts, to recover all measurable information in the Bernoulli system. The sigma-algebra $\mathcal{F}$ sees the coordinates $0,1,2,\dots$. Applying $\sigma$ once moves this window one step to the left, so $\sigma\mathcal{F}$ sees the coordinates $-1,0,1,\dots$. Applying $\sigma^m$ sees the coordinates $-m,-m+1,\dots$. Formally, for every $m \in \mathbb{N}_0$, \begin{align*} \sigma^m\mathcal{F} = \sigma(X_k : k \ge -m). \end{align*} Now fix any $j \in \mathbb{Z}$. If $j \ge 0$, then $X_j$ is already measurable with respect to $\mathcal{F}$. If $j < 0$, choose $m := -j$; then $j \ge -m$, so $X_j$ is measurable with respect to $\sigma^m\mathcal{F}$. Hence every coordinate map $X_j$ is measurable with respect to $\bigvee_{m=0}^{\infty}\sigma^m\mathcal{F}$. Conversely, every $\sigma^m\mathcal{F}$ is a sub-sigma-algebra of $\mathcal{B}$, so their join is contained in $\mathcal{B}$. Since the product sigma-algebra is generated by all coordinate maps, we conclude \begin{align*} \bigvee_{m=0}^{\infty} \sigma^m\mathcal{F} = \sigma(X_k : k \in \mathbb{Z}) = \mathcal{B}. \end{align*}[/guided]

[step:Identify the decreasing intersection with the future tail sigma-algebra] For each $n \in \mathbb{N}_0$, the shifted preimage satisfies \begin{align*} \sigma^{-n}\mathcal{F} = \sigma(X_k : k \ge n). \end{align*} Therefore \begin{align*} \bigcap_{n=0}^{\infty} \sigma^{-n}\mathcal{F} = \bigcap_{n=0}^{\infty} \sigma(X_k : k \ge n). \end{align*} Denote this tail sigma-algebra by \begin{align*} \mathcal{T}_+ := \bigcap_{n=0}^{\infty} \sigma(X_k : k \ge n). \end{align*} It remains to prove that $\mathcal{T}_+$ is trivial modulo $\mu$. [/step]

[step:Prove that the future tail sigma-algebra is trivial][claim:Use the pi-lambda principle for generated sigma-algebras] Let $\mathcal{P}$ be a pi-system on a set $S$, and let $\mathcal{L}$ be a lambda-system on $S$ with $\mathcal{P}\subset\mathcal{L}$. Then $\sigma(\mathcal{P})\subset\mathcal{L}$. [/claim] [proof] Let $\mathcal{D}$ be the intersection of all lambda-systems on $S$ that contain $\mathcal{P}$. Then $\mathcal{D}$ is a lambda-system and $\mathcal{P}\subset\mathcal{D}$. For each $P\in\mathcal{P}$, the class \begin{align*} \mathcal{D}_P:=\{D\in\mathcal{D}:D\cap P\in\mathcal{D}\} \end{align*} is a lambda-system containing $\mathcal{P}$, because $\mathcal{P}$ is closed under finite intersections. Hence $\mathcal{D}_P=\mathcal{D}$ for every $P\in\mathcal{P}$. For each $D\in\mathcal{D}$, the class \begin{align*} \mathcal{E}_D:=\{E\in\mathcal{D}:E\cap D\in\mathcal{D}\} \end{align*} is again a lambda-system containing $\mathcal{P}$, so $\mathcal{E}_D=\mathcal{D}$. Therefore $\mathcal{D}$ is closed under finite intersections. Since a lambda-system closed under finite intersections is a sigma-algebra, $\mathcal{D}$ is a sigma-algebra containing $\mathcal{P}$, so $\sigma(\mathcal{P})\subset\mathcal{D}\subset\mathcal{L}$. [/proof] Let \begin{align*} \mathcal{G}_n := \sigma(X_k : 0 \le k < n) \end{align*} for $n \in \mathbb{N}$, and let \begin{align*} \mathcal{G}_\infty := \sigma(X_k : k \ge 0). \end{align*} For each $n \in \mathbb{N}$, the coordinate family $(X_k)_{0 \le k < n}$ is independent of the coordinate family $(X_k)_{k \ge n}$ under the product measure $\mu$. We justify the passage from coordinate independence to sigma-algebra independence. Let $\mathcal{P}_n$ denote the pi-system of finite-coordinate cylinder sets in $\mathcal{G}_n$, and let $\mathcal{Q}_n$ denote the pi-system of finite-coordinate cylinder sets in $\sigma(X_k : k \ge n)$. Product structure gives \begin{align*} \mu(P \cap Q) = \mu(P)\mu(Q) \end{align*} for every $P \in \mathcal{P}_n$ and every $Q \in \mathcal{Q}_n$. Fixing $P \in \mathcal{P}_n$ and applying the pi-lambda claim just proved to the lambda-system of all $Q \in \sigma(X_k : k \ge n)$ satisfying this identity extends the identity from $\mathcal{Q}_n$ to $\sigma(X_k : k \ge n)$. Then fixing $Q \in \sigma(X_k : k \ge n)$ and applying the same claim to the lambda-system of all $P \in \mathcal{G}_n$ satisfying the identity extends it from $\mathcal{P}_n$ to $\mathcal{G}_n$. Hence $\mathcal{G}_n$ is independent of $\sigma(X_k : k \ge n)$. If $E \in \mathcal{T}_+$, then $E \in \sigma(X_k : k \ge n)$ for every $n \in \mathbb{N}$, so $E$ is independent of $\mathcal{G}_n$ for every $n$. We now pass from independence of each finite block to independence of $\mathcal{G}_\infty$. For the fixed set $E$, define \begin{align*} \mathcal{L}_E := \{B \in \mathcal{G}_\infty : \mu(E \cap B) = \mu(E)\mu(B)\}. \end{align*} The class $\mathcal{L}_E$ is a lambda-system: it contains $X$, it is closed under complements inside $X$ by subtracting the identity from $\mu(E)$, and it is closed under countable disjoint unions by countable additivity of $\mu$. The class $\bigcup_{n=1}^{\infty}\mathcal{G}_n$ is a pi-system, it generates $\mathcal{G}_\infty$, and it is contained in $\mathcal{L}_E$ because $E$ is independent of each $\mathcal{G}_n$. By the pi-lambda claim proved above, $\mathcal{L}_E = \mathcal{G}_\infty$. Hence $E$ is independent of $\mathcal{G}_\infty$. This is the Kolmogorov zero-one mechanism for independent tails, written here directly from the proved pi-lambda claim. But $E \in \mathcal{T}_+ \subset \mathcal{G}_\infty$, so $E$ is independent of itself. Thus \begin{align*} \mu(E) = \mu(E \cap E) = \mu(E)^2. \end{align*} Therefore $\mu(E) \in \{0,1\}$. Hence $\mathcal{T}_+$ is trivial modulo $\mu$.[/step]

[guided]The key point is that a future-tail event cannot depend on any finite initial block of future coordinates. We make this precise using independence and the pi-lambda theorem. For $n \in \mathbb{N}$, define \begin{align*} \mathcal{G}_n := \sigma(X_k : 0 \le k < n) \end{align*} and define \begin{align*} \mathcal{G}_\infty := \sigma(X_k : k \ge 0). \end{align*} The product measure $\mu$ makes disjoint coordinate blocks independent. To pass from independence of finite cylinder events to independence of the sigma-algebras they generate, let $\mathcal{P}_n$ be the pi-system of finite-coordinate cylinder sets in $\mathcal{G}_n$, and let $\mathcal{Q}_n$ be the pi-system of finite-coordinate cylinder sets in $\sigma(X_k : k \ge n)$. Product structure gives \begin{align*} \mu(P \cap Q) = \mu(P)\mu(Q) \end{align*} for every $P \in \mathcal{P}_n$ and $Q \in \mathcal{Q}_n$. Fixing $P$ and applying the pi-lambda claim proved in the exact part of this step extends this identity to every $Q \in \sigma(X_k : k \ge n)$. Fixing such a $Q$ and applying the same claim again extends the identity to every $P \in \mathcal{G}_n$. Thus $\mathcal{G}_n$ is independent of $\sigma(X_k : k \ge n)$. Now let $E \in \mathcal{T}_+$. Since \begin{align*} \mathcal{T}_+ = \bigcap_{n=0}^{\infty}\sigma(X_k : k \ge n), \end{align*} the event $E$ belongs to $\sigma(X_k : k \ge n)$ for every $n \in \mathbb{N}$. Therefore $E$ is independent of $\mathcal{G}_n$ for every $n \in \mathbb{N}$. We must upgrade independence from every finite block $\mathcal{G}_n$ to the full future sigma-algebra $\mathcal{G}_\infty$. Define \begin{align*} \mathcal{L}_E := \{B \in \mathcal{G}_\infty : \mu(E \cap B) = \mu(E)\mu(B)\}. \end{align*} This class is a lambda-system: it contains $X$, it is closed under complements inside $X$ because the defining identity can be subtracted from $\mu(E)$, and it is closed under countable disjoint unions by countable additivity of $\mu$. The class $\bigcup_{n=1}^{\infty}\mathcal{G}_n$ is a pi-system, generates $\mathcal{G}_\infty$, and is contained in $\mathcal{L}_E$ because $E$ is independent of every $\mathcal{G}_n$. The pi-lambda claim gives $\mathcal{L}_E = \mathcal{G}_\infty$, so $E$ is independent of $\mathcal{G}_\infty$. This is exactly the usual Kolmogorov zero-one mechanism for independent tails, but the argument above proves the needed special case directly. Finally, $E \in \mathcal{T}_+ \subset \mathcal{G}_\infty$, so $E$ is independent of itself. Hence \begin{align*} \mu(E) = \mu(E \cap E) = \mu(E)^2. \end{align*} The only numbers in $[0,1]$ satisfying $a = a^2$ are $0$ and $1$, so $\mu(E) \in \{0,1\}$. Since every $E \in \mathcal{T}_+$ has measure $0$ or $1$, the future tail sigma-algebra $\mathcal{T}_+$ is trivial modulo $\mu$.[/guided]

[step:Conclude the K-property for the Bernoulli shift] The sigma-algebra $\mathcal{F}$ satisfies \begin{align*} \sigma^{-1}\mathcal{F} \subset \mathcal{F}, \end{align*} its forward translates generate the whole sigma-algebra, \begin{align*} \bigvee_{m=0}^{\infty}\sigma^m\mathcal{F} = \mathcal{B}, \end{align*} and its decreasing preimage tail is trivial, \begin{align*} \bigcap_{n=0}^{\infty}\sigma^{-n}\mathcal{F} \end{align*} is the null sigma-algebra modulo $\mu$. These are exactly the defining conditions for $(X,\mathcal{B},\mu,\sigma)$ to have the K-property. [/step]

[step:Transfer the property across measure-theoretic isomorphism] Let $(Y,\mathcal{C},\lambda,T)$ be measure-theoretically isomorphic to $(X,\mathcal{B},\mu,\sigma)$. Let $\Phi: Y \to X$ be an invertible measure-space isomorphism modulo null sets such that $\Phi \circ T = \sigma \circ \Phi$ modulo $\lambda$-null sets. Choose full-measure sets $Y_0 \in \mathcal{C}$ and $X_0 \in \mathcal{B}$ such that $\Phi|_{Y_0}: Y_0 \to X_0$ is a bimeasurable measure-space isomorphism and the intertwining relation holds on $Y_0 \cap T^{-1}(Y_0)$ after replacing $Y_0$ by a further full-measure invariant representative if necessary. In this step, equalities of sigma-algebras mean equality after completion modulo null sets. Define \begin{align*} \mathcal{H} := \Phi^{-1}\mathcal{F} = \{\Phi^{-1}(A) : A \in \mathcal{F}\}. \end{align*} Then \begin{align*} T^{-1}\mathcal{H} = T^{-1}\Phi^{-1}\mathcal{F} = \Phi^{-1}\sigma^{-1}\mathcal{F} \subset \Phi^{-1}\mathcal{F} = \mathcal{H}, \end{align*} where the equality follows because $T^{-1}(\Phi^{-1}(A))$ and $\Phi^{-1}(\sigma^{-1}(A))$ differ by a $\lambda$-null set for every $A \in \mathcal{F}$, using the intertwining relation $\Phi \circ T = \sigma \circ \Phi$ modulo $\lambda$-null sets. Similarly, pullback by $\Phi$ preserves countable joins after completion. Indeed, on the full-measure representatives $Y_0$ and $X_0$, the map $\Phi|_{Y_0}$ is an actual bimeasurable isomorphism, so inverse image commutes with the sigma-algebra operation generated by a countable family. Passing back to completions only adds null sets. Hence \begin{align*} \bigvee_{m=0}^{\infty} T^m\mathcal{H} = \Phi^{-1}\left(\bigvee_{m=0}^{\infty}\sigma^m\mathcal{F}\right) = \Phi^{-1}\mathcal{B} = \mathcal{C} \end{align*} modulo $\lambda$-null sets. We spell out the corresponding statement for countable intersections in the completed measure algebras. Let $\widehat{\mathcal{B}}$ and $\widehat{\mathcal{C}}$ denote the completions modulo null sets, and let \begin{align*} \Phi^*:\widehat{\mathcal{B}}\to\widehat{\mathcal{C}} \end{align*} be the Boolean algebra isomorphism induced by inverse image under $\Phi$. Its inverse is induced by the measurable inverse on the full-measure representatives $X_0$ and $Y_0$. If $D\in\bigcap_{n=0}^{\infty}T^{-n}\mathcal{H}$ modulo null sets, define $A:=(\Phi^*)^{-1}(D)$ in $\widehat{\mathcal{B}}$. Since $D\in T^{-n}\mathcal{H}$ for every $n$, the intertwining relation gives $A\in\sigma^{-n}\mathcal{F}$ in $\widehat{\mathcal{B}}$ for every $n$. Hence $A\in\bigcap_{n=0}^{\infty}\sigma^{-n}\mathcal{F}$ modulo $\mu$-null sets, and $D=\Phi^{-1}(A)$ modulo $\lambda$-null sets. The reverse inclusion follows because inverse images preserve membership in each $\sigma^{-n}\mathcal{F}$. Therefore \begin{align*} \bigcap_{n=0}^{\infty} T^{-n}\mathcal{H} = \Phi^{-1}\left(\bigcap_{n=0}^{\infty}\sigma^{-n}\mathcal{F}\right) \end{align*} modulo $\lambda$-null sets. The right-hand side is trivial modulo $\lambda$, because $\Phi$ sends $\lambda$-null sets to $\mu$-null sets and pulls back $\mu$-conull sets to $\lambda$-conull sets. Thus $(Y,\mathcal{C},\lambda,T)$ has the K-property. Consequently every Bernoulli system isomorphic to the two-sided Bernoulli shift has the K-property. [/step]