Let $(\Omega, \mathcal{F}, \mathbf{P})$ be a probability space and $\{X_\lambda\}_{\lambda \in \Lambda}$ a (possibly uncountable) family of real random variables, which are only defined up to null-sets, i.e. in fact $X_\lambda$ $(\lambda \in \Lambda)$ is an equivalence class with respect to almost sure equality. Then the essential supremum of $\{X_\lambda\}_{\lambda \in \Lambda}$ is defined as the unique (up to null-sets) random variable $Y$ such that
- $Y \geq X_\lambda$ almost sure for every $\lambda \in \Lambda$,
- Whenever a random variable $\tilde{Y}$ satisfies (1) it is $Y \leq \tilde{Y}$ almost sure.
At several points, I have seen two motivations for this definition: (1) The essential supremum is measurable even if $\Lambda$ is uncountable, while the same does not hold for the "normal" pointwise supremum. (2) The "normal" supremum might not reflect the intuitive meaning of a supremum.
However, I would have expected another motivation, namely: The "normal" pointwise supremum is not well-defined if $\Lambda$ is uncountable and the $X_\lambda$ are only defined up to null sets.
Does this argument hold? Is it true that the "normal" supremum would not be well-defined?
Best Answer
You can't really take a pointwise supremum of a set of equivalence classes of random variables, since there's no such thing as $X_\lambda(\omega)$ for any $\omega\in\Omega$. $X_\lambda$ is not a map, so $X_\lambda(\omega)$ is meaningless, and so is $\sup_{\lambda\in\Lambda}X_\lambda(\omega)$. So in that sense, yes, the pointwise supremum is not well-defined in your case.
But this has nothing to do with wether $\Lambda$ is finite, countable, or uncountable. In fact, if $X_\lambda$ were proper random variables and not equivalence classes of such, then the pointwise supremum would be well-defined. That's one of the main motivations to construct the reals: every subset of the reals is supposed to have a supremum, and $\{X_\lambda(\omega)~\vert~\lambda\in\Lambda\}$ is certainly a subset of the reals for any $\omega\in\Omega$. The supremum may no longer be real valued, since it can be $\infty$, but the same is true for the essential supremum, so that's not a good motivation for the essential supremum.