The upper modified box-counting dimension of $F \subseteq{\mathbb{R}^n}$ is defined as
$\overline{\dim}_{MB}F = \inf \left\{ \sup_{i} \overline{\dim}_{B}F_i; F \subseteq \bigcup_{i=1}^{\infty} F_i\right\}$, where the $F_i$ are nonempty bounded subsets of $F$ and the infimum is over all countable or finite covers $\{F_i\}$ of $F$, and $\overline{\dim}_{B}F_i$ is the box-counting (Minkowski) dimension of $F_i$.
Analogue for the lower modified box-counting dimenion: $\underline{\dim}_{MB}F = \inf \left\{ \sup_{i} \underline{\dim}_{B}F_i; F \subseteq \bigcup_{i=1}^{\infty} F_i\right\}$.
I need to prove that the modified box-counting dimension is countably stable: $\overline{\dim}_{MB}\left( \bigcup_{i=1}^{\infty} F_i\right) = \sup_i \overline{\dim}_{MB}(F_i)$, and analogue for the lower box-counting dimension.
I noticed that for a given $1 \le j \le \infty$, since $F_j \subseteq \bigcup_{i=1}^{\infty}F_i$, then by monotonicity (already proven), $\overline{\dim}_{MB}F_j \le \overline{\dim}_{MB}\left( \bigcup_{i=1}^{\infty} F_i \right)$, hence the RHS is an upper bound, so by the definition of supremum I have $\overline{\dim}_{MB}\left( \bigcup_{i=1}^{\infty} F_i\right) \ge \sup_j \overline{\dim}_{MB}(F_j) = \sup_i \overline{\dim}_{MB}(F_i) $
But I'm having trouble proving that $\overline{\dim}_{MB}\left( \bigcup_{i=1}^{\infty} F_i\right) \le \sup_i \overline{\dim}_{MB}(F_i)$, how can I do this?
Thanks.