Consider a sequence of stochastic processes $$((X_f^{(n)})_{f \in F})_{n \in \mathbb{N}}.$$ All the random variables $X_f^{(n)}$ are defined on the same probability space and assume only non-negative values. In my case $F$ is a unit ball of some separable Banach space. I am looking for a way to express the condition \begin{equation}\label{condition} \lim_{n \rightarrow \infty} \sup_{f \in F} \mathbb{E}\left[ X_{f}^{(n)} \right] = 0 \end{equation} in a fashion similar to ``convergence in probability + uniform integrability = convergence in $L^1$''. It would be easy if either there was no supremum over $F$ or the supremum was inside of the expectation, but in the above case I cannot really come up with an if-and-only-if condition.
Any help would be greatly appreciated!
Consider the conditions $$ \tag{1}\lim_{R\to\infty}\limsup_{n\to\infty}\sup_{f\in F}\mathbb E\left[X_f^{(n)}\mathbf{1}\{X_f^{(n)}>R\}\right]=0 $$ and $$ \tag{2}\forall \varepsilon >0,\quad\lim_{n\to\infty}\sup_{f\in F}\mathbb P\left(X_f^{(n)}>\varepsilon\right)=0. $$ The decomposition
$$ \mathbb E\left[X_f^{(n)} \right]=\mathbb E\left[X_f^{(n)}\mathbf{1}\{X_f^{(n)}\leqslant \varepsilon\}\right]+\mathbb E\left[X_f^{(n)}\mathbf{1}\{\varepsilon <X_f^{(n)}\leqslant R\}\right]+\mathbb E\left[X_f^{(n)}\mathbf{1}\{X_f^{(n)}>R\}\right] $$ (valid for $\varepsilon< R$) gives the bound $$ \mathbb E\left[X_f^{(n)} \right]\leqslant \varepsilon +R\mathbb{P}\left(X_f^{(n)}>\varepsilon\right) +\mathbb E\left[X_f^{(n)}\mathbf{1}\{X_f^{(n)}>R\}\right]. $$ Consequently, for each $\varepsilon< R$, $$ \limsup_{n\to\infty}\sup_{f\in F}\mathbb E\left[X_f^{(n)} \right]\leqslant \varepsilon +R\limsup_{n\to\infty}\sup_{f\in F}\mathbb{P}\left(X_f^{(n)}>\varepsilon\right) +\limsup_{n\to\infty}\sup_{f\in F}\mathbb E\left[X_f^{(n)}\mathbf{1}\{X_f^{(n)}>R\}\right]. $$ By (2), this reduces to $$ \limsup_{n\to\infty}\sup_{f\in F}\mathbb E\left[X_f^{(n)} \right]\leqslant \varepsilon +\limsup_{n\to\infty}\sup_{f\in F}\mathbb E\left[X_f^{(n)}\mathbf{1}\{X_f^{(n)}>R\}\right]. $$ and by (1), we get that $\limsup_{n\to\infty}\sup_{f\in F}\mathbb E\left[X_f^{(n)} \right]\leqslant \varepsilon$, which is sufficient to conclude that $$\tag{3}\lim_{n\to\infty}\sup_{f\in F}\mathbb E\left[X_f^{(n)} \right]=0.$$
Actually, the combination of conditions (1) and(2) is equivalent to (3), but might be easier to check in the practise.