In a game, if $H_n$ is the money that we bet, why is $H_n$ previsible (i.e. $\mathcal F_{n-1}$ measurable) ?
For example, suppose we play with a dice. If the number is $n\in\{1,3,6\}$, then we loose $n-$time the money we bet. If $n\in \{2,4,5\}$, we win $n$ time the money we bet. Let $H_n$ the money we bet at the $n$-th thrown. Why is $H_n$ previsible ? The fact that the gain are $\mathcal F_n$ measurable, is clear. But I don't understand why the bet is $\mathcal F_{n-1}$ measurable
The complete details of the general situation you have in mind isn't clear to me, but I'm assuming that $\ \left\{\mathcal{F}_n\right\} $ is a filtration, (i.e. $ \mathcal{F}_{n-1}\subseteq\mathcal{F}_n$ for all $\ n\ $). In situations like those you appear to be describing, $ \mathcal{F}_{n-1} $ would typically be the set of events whose occurrence or non-occurrence is already determinable by the $n^\text{th} $ instant, and thus represents all the information available to you when you decide what to choose for $ H_n $. If this is so, then the $ H_n $ you choose must be $ \mathcal{F}_{n-1}$-measurable, because no relevant information other than that represented by $ \mathcal{F}_{n-1} $ is available to you when you make your choice.
In your example, you would typically take the sample space to be the set of sequences, $$ \Omega = \left\{\,\left\{d_i\right\}_{i=1}^\infty\big\vert\, d_i\in\{1,2,3,4,5,6\} \text{ for all } i\right\}\ , $$ and $ \mathcal{F}_n $ to be the power set of the set of all $ 6^n $ $ n$-long cylinders, each of which comprises all sequences in which the first $n $ terms are fixed at some specific $\ n$-long sequence of values. Making $H_n $ $ \mathcal{F}_{n-1}$-measurable here means that it can be made a function of the outcomes of the first $ n-1 $ tosses of the die, but cannot depend on the outcomes of any tosses beyond those.