 Solution

Let $A_n={\sum }_{i=1}^{m_n}f(X_i)$ and $T_n={\sum }_{i=m_n+1}^{n}f(X_i)$. Since $|A_n|\le B{m}_n$, $\mathrm{Var}(A_n)\le B^2{m}_n^2$. The Cauchy-Schwarz inequality applied to the centered sums givesDivide by $\sqrt{n}$. The first term is at most $B{m}_n/\sqrt{n}\to 0$, while part (b) and $(n-m_n)/n\to 1$ give a limsup of at most $\sqrt{\gamma {\mathrm{Var}}_{\mu }(f)}$ for the second. Squaring provesThis argument also covers ${\mathrm{Var}}_{\mu }(f)=0$; it does not require independence between the discarded prefix and the retained path. The same chain-dependent $\gamma$ as in part (b) suffices.