I' m a worker and I'm self studying stochastic calculus. I'm struggling with the definition of bounded process. It seems to me that each author uses different definition.
Now, in the book I'm studying, there is no a specific definition of bounded process but the author says:
If $\Pi$ is some property of a sample path, then we say the process has property $\Pi$ if every sample path $t\mapsto X_t(\omega)$ has property “$\Pi$”.
So a bounded process $X$, that takes value in $\mathbb{R}$, on the measurable space $(\Omega,\mathcal{F})$ should be a proces such that for each $\omega \in\Omega $ there exists a finite constant $M_{\omega}$ such that: $$ |X_{t}(\omega)|< M_{\omega} \; \; \; \forall t \in \mathcal{T};$$ where $\mathcal{T}$ is an index set that can be countable or not.\
My question is: can I conclude that there exists a finite constant $M$ such that: $$| X_{t}(\omega)|< M \; \; \; \forall t \in \mathcal{T} \; \forall \omega \in \Omega$$ defining: $$ M=\sup_{\omega \in \Omega} M_{\omega}$$.
In the title I called this property uniformly boundedness.
In general, without further assumptions, it is not possible to conclude uniform boundedness. For example, let $X_t(\omega)=Y(\omega)$ where $Y$ is a non bounded random variable (that is, $Y\notin \mathbb L^\infty$ or $\mathbb P(\lVert Y\rVert\gt n)>0$ for all $n$). For example, if $\Omega$ is $(0,1)$ endowed with the Lebesgue measure, take $Y\colon\omega\mapsto 1/\omega$.