A measurable observation does not increase stationary mixing coefficients

For any finite indices $k_1,\dots ,k_m$ and time shift $t\ge 0$, strict stationarity [F1] makes $(X_{k_1+t},\dots ,X_{k_m+t})$ equal in distribution to $(X_{k_1},\dots ,X_{k_m})$. Applying the measurable map $h$ coordinatewise preserves equality in distribution, so $(Y_{k_1+t},\dots ,Y_{k_m+t})$ has the same law as $(Y_{k_1},\dots ,Y_{k_m})$. Thus $Y$ is strictly stationary.

For every $j$, measurability gives $\sigma (Y_0,\dots ,Y_j)\subseteq \sigma (X_0,\dots ,X_j)$ and $\sigma (Y_{j+n},Y_{j+n+1},\dots )\subseteq \sigma (X_{j+n},X_{j+n+1},\dots )$. Therefore every event pair used in the suprema for ${\alpha }_Y$ and ${\varphi }_Y$ is also allowed for $X$, with the same probability difference. Taking the suprema in [F2] proves ${\alpha }_Y(n)\le {\alpha }_X(n)$ and ${\varphi }_Y(n)\le {\varphi }_X(n)$.

Every square-integrable past variable and future variable measurable with respect to the corresponding $Y$-sigma-algebras is also measurable with respect to the corresponding $X$-sigma-algebras. Its correlation is unchanged, so the supremum for ${\rho }_Y(n)$ is over a subset of the pairs allowed for ${\rho }_X(n)$. Hence ${\rho }_Y(n)\le {\rho }_X(n)$.