[proofplan] We decompose $L^2$ of the product space into tensor-coordinate subspaces. Each finite non-empty coordinate shape has all of its translates under the shift, and those translates are mutually orthogonal. Choosing one representative of each translation class gives countably many bilateral-shift summands. The bilateral shift has Lebesgue spectral measure, so the spectrum is countable Lebesgue. Strong mixing follows from the Rajchman spectral criterion. [/proofplan] [step:Decompose the product $L^2$ space] Let \begin{align*} K:=L^2(A,p;\mathbb{C}), \qquad K_0:=\left\{h\in K:\sum_{a\in A}h(a)p_a=0\right\}. \end{align*} Then \begin{align*} K=\mathbb{C}\mathbf{1}\oplus K_0. \end{align*} The finite-coordinate cylinder functions are dense in $L^2(A^{\mathbb{Z}},\mu)$. Every such function decomposes into a finite sum of tensor terms supported on finite sets $F\subset\mathbb{Z}$, where the coordinates in $F$ lie in copies of $K_0$ and the remaining coordinates are constants. Therefore \begin{align*} L^2_0(A^{\mathbb{Z}},\mu) = \bigoplus_{\varnothing\neq F\subset\mathbb{Z}\text{ finite}} H_F, \end{align*} where $H_F$ is the finite-dimensional tensor subspace with nonconstant factors exactly on $F$. [/step] [step:Group finite coordinate sets into shift orbits] The Koopman operator $U_\sigma f=f\circ\sigma$ maps $H_F$ unitarily onto $H_{F+1}$, where \begin{align*} F+1:=\{n+1:n\in F\}. \end{align*} Choose one representative $F$ from each translation class of non-empty finite subsets of $\mathbb{Z}$, for instance by requiring $\min F=0$. There are countably many such representatives. For each representative $F$, choose an orthonormal basis $(e_{F,j})_{j=1}^{d_F}$ of $H_F$. For fixed $F$ and $j$, the vectors \begin{align*} U_\sigma^n e_{F,j}, \qquad n\in\mathbb{Z}, \end{align*} are orthonormal, because they lie in mutually orthogonal coordinate subspaces $H_{F+n}$. Their closed span is unitarily equivalent to $\ell^2(\mathbb{Z})$, with $U_\sigma$ corresponding to the bilateral shift. [/step] [step:Identify the spectral type] The bilateral shift on $\ell^2(\mathbb{Z})$ is unitarily equivalent, by Fourier series, to multiplication by $z$ on $L^2(\mathbb{T},m_{\mathbb{T}})$, where $m_{\mathbb{T}}$ is Haar probability measure on the unit circle. Hence each cyclic summand above has Lebesgue spectral type. There are only countably many representatives $F$ and finitely many basis vectors for each $H_F$. Therefore $L^2_0$ is an orthogonal direct sum of countably many Lebesgue spectral summands. This is countable Lebesgue spectrum. [/step] [step:Deduce strong mixing] Haar measure on $\mathbb{T}$ is Rajchman by the [Riemann Lebesgue Lemma](/theorems/526), and countable direct sums of Lebesgue spectral measures remain absolutely continuous with respect to Haar measure on each cyclic component. Hence every zero-mean spectral measure is Rajchman. By the [Spectral Characterisation of Strong Mixing](/theorems/3456), the Bernoulli shift is strongly mixing. [/step]