Imagine I have a given sample $X_1, \ldots, X_T$ of some process; Also the realisations are identically distributed and we denote with $\mathbb E(X_1)=\mu$
Imagine I bootstrap it the standard way, so choose $T$ realisations $X_{i_1},\ldots,X_{i_T}$ randomly from the given sample with replacement; Denote the bootstrap sample as $X_1^*,\ldots,X_T^*$ and denote with $\bar{X}_n$ the mean of the original sample and $\bar{X}_n^*$ the mean of the bootstrap sample;
Now I know that:
$$\sup_x\left\vert \mathbb P\left(\sqrt{T}(\bar{X}_n^*-\bar{X}_n)\leq x \mid X_1,\ldots,X_T\right)-\mathbb P\left(\sqrt{T}(\bar{X}_n-\mu)\leq x\right) \right \vert \to 0$$ almost surely as $T$ goes to infinity;
Now I want to show that:
$$\left\vert\operatorname{Var}^*\bigg(\sqrt{T}(\bar{X}_n^*-\bar{X}_n)\bigg)-\operatorname{Var}\bigg(\sqrt{T}(\bar{X}_n-\mu)\bigg) \right \vert\to 0$$ almost surely as $T$ goes to infinity;
here $\operatorname{Var}^*(\cdot)$ denotes the variance operator of the resampling distribution $P^∗$ conditional on our original sample $X_1,\ldots,X_T$.
We see it is very similar- so it is mostly only about turning the supremum of probabilities into a variance;
But how can I do it?