Let μi,νi\mu_i, \nu_i be probability measures on the finite space (Ωi,2Ωi)(\Omega_i, 2^{\Omega_i} ), i∈[n]i \in [n]. Define μ=∏i∈[n]μi\mu = \prod_{i \in [n]} \mu_i and ν=∏i∈[n]νi\nu = \prod_{i \in [n]} \nu_i be the product measure on the product space (Ω,2Ω)(\Omega, 2^{\Omega}). Then

∥μ−ν∥TV≤∑i∈[n]∥μi−νi∥TV, \Vert \mu - \nu \Vert_{TV} \le \sum_{i \in [n] } \Vert \mu_i - \nu_i \Vert_{TV},

where ∥⋅∥TV\Vert \cdot \Vert_{TV} is the total variation distance.

Proof. Consider an x=(x1,…,xn)∈Ωx = (x_1, \ldots, x_n) \in \Omega.

For each i∈0,…,ni \in 0, \ldots, n, let

s(i)≐μ1(x1)⋯μi(xi)⋅νi+1(xi+1)⋯νn(xn). s(i) \doteq \mu_1 (x_1) \cdots \mu_i(x_i) \cdot \nu_{i + 1} (x_{i + 1} ) \cdots \nu_n (x_n).

So s(0)=ν(x)s(0) = \nu(x) and s(n)=μ(x)s(n) = \mu(x). It follows that

ν(x)−μ(x)=∑i∈[n](s(i−1)−s(i))=∑i∈[n]((νi(xi)−μi(xi))⋅μ1(x1)⋯μi−1(xi−1)⋅νi+1(xi+1)⋯νn(xn)). \begin{aligned} \nu(x) - \mu(x) &= \sum_{i \in [n]} \left( s(i - 1) - s(i) \right) \\ &= \sum_{i \in [n]} \left( \big( \nu_i(x_i) - \mu_i(x_i) \big) \cdot \mu_1 (x_1) \cdots \mu_{i - 1} (x_{i - 1}) \cdot \nu_{i + 1} (x_{i + 1} ) \cdots \nu_n (x_n)\right). \end{aligned}

Hence,

2⋅∥μ−ν∥TV=∑x∈Ω∣ν(x)−μ(x)∣=∑x∈Ω∣∑i∈[n](νi(xi)−μi(xi))⋅μ1(x1)⋯μi−1(xi−1)⋅νi+1(xi+1)⋯νn(xn)∣≤∑x∈Ω∑i∈[n]∣νi(xi)−μi(xi)∣⋅μ1(x1)⋯μi−1(xi−1)⋅νi+1(xi+1)⋯νn(xn)=∑i∈[n]∑xi∈Ωi∣νi(xi)−μi(xi)∣=∑i∈[n]2⋅∥μi−νi∥TV. \begin{aligned} 2 \cdot \Vert \mu - \nu \Vert_{TV} &= \sum_{x \in \Omega} \vert \nu(x) - \mu(x) \vert \\ &= \sum_{x \in \Omega} \left\vert \sum_{i \in [n]} \big( \nu_i(x_i) - \mu_i(x_i) \big) \cdot \mu_1 (x_1) \cdots \mu_{i - 1} (x_{i - 1}) \cdot \nu_{i + 1} (x_{i + 1} ) \cdots \nu_n (x_n)\right\vert \\ &\le \sum_{x \in \Omega} \sum_{i \in [n]} \left\vert \nu_i(x_i) - \mu_i(x_i) \right\vert \cdot \mu_1 (x_1) \cdots \mu_{i - 1} (x_{i - 1}) \cdot \nu_{i + 1} (x_{i + 1} ) \cdots \nu_n (x_n) \\ &= \sum_{i \in [n]} \sum_{x_i \in \Omega_i} \left\vert \nu_i(x_i) - \mu_i(x_i) \right\vert \\ &= \sum_{i \in [n]} 2 \cdot \Vert \mu_i - \nu_i \Vert_{TV}. \end{aligned}

□\square