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Let $(S,\mathcal{S},P)$ be a probability space. Suppose that $\mathcal{F}$ is a countable set of real measurable functions on $S$ such that $\int fdP=0$ and $\sup_{f\in\mathcal{F}}\|f\|_\infty\leq 1$. Let $X_1,\dots,X_n$ be random variables taking values in $S$ and define $$Z = \sup_{f\in\mathcal{F}}\sum_{i=1}^n f(X_i),\qquad Z_k = \sup_{f\in\mathcal{F}}\sum_{i\ne k}f(X_i),\qquad k=1,\dots,n.$$

We need to prove that $\sum_{k=1}^n(Z - Z_k)\leq Z$. The prof goes as follows. Assume that $\mathcal{F}$ is finite, then there exists $f\in\mathcal{F}$ such that $$(n-1)Z = (n-1)\sum_{i=1}^nf(X_i) = \sum_{k=1}^n\sum_{i\ne k}f(X_i)\leq \sum_{k=1}^nZ_k,$$ which gives the result. Now the paper says that the case when $\mathcal{F}$ is not countable and so the supremum is not achievable, can be established by considering a sequence of finite sets and passing to the limit.

I'm looking for justification for this, why can we claim that there exists a sequence of sets such that $$\lim_{m\to\infty}\sup_{f\in\mathcal{F}_m}\sum_{k=1}^nf(X_k) = \sup_{f\in\mathcal{F}}\sum_{k=1}^nf(X_k)?$$

This relates to another my question Is countable pointwise supremum of bounded functions achievable?, namely the comment given by Brian M. Scott.

This is a part of the proof of the Bousquet inequality appeared in Bousquet, "Concentration Inequalities for Sub-Additive Functions Using the Entropy Method", 2003.

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Given $\varepsilon>0$, there exists $f$ such that $Z<\sum_{i=1}^n f(X_i)+\varepsilon$. Hence, $$ \begin{split} (n-1)Z &= (n-1)\left(\sum_{k=1}^nf(X_k)+\varepsilon\right) \\ &= \sum_{k=1}^n\sum_{i\ne k}f(X_i)+(n-1)\varepsilon\leq \sum_{k=1}^nZ_k+(n-1)\varepsilon. \end{split} $$ In other words, there is no need to consider several cases.

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