[I realize that this question may be too basic and likely to contain a fundamental misunderstanding, but similar questions such as this one on the site are beyond my math level. So please assume very limited understanding of measure theory. If at all possible an answer in plain English would be ideal.]
A probability space is defined as the triple ($\Omega,\mathcal{F},P$).
The sample space and the sigma-algebra of events of a probability space -($\Omega,\mathcal{F}$)- would seem to fulfill the conditions of a topological space, ($X,\tau$):
- There is always a null event ($P(\{\emptyset\} = 0)$) and a certain event ($P(\{\Omega\})=1$). This corresponds to the condition, "The empty set and $X$ itself belong to $\tau$" in a topology sapce.
- Any union of events belongs to $\mathcal{F}$ (closed under countable unions). The counterpart in topology spaces: "Any (finite or infinite) union of members of $\tau$ still belongs to $\tau$."
- The intersection of any number of events belongs to $\mathcal{F}$ (closed under countable intersections). In topology spaces: "The intersection of any finite number of members of $\tau$ still belongs to $\tau$."
Can we then say that the sample space $\Omega$ and the $\sigma$-algebra, ($\Omega, \mathcal{F}$), form a topology on $\Omega$?
I believe that ($\Omega, \mathcal{F}$) also fulfills the conditions of a measurable space, so I may be indirectly asking the difference between a topological and a measurable space.
Is the coincidence of criteria an accident among conceptually disparate mathematical objects: probability (or measurable) spaces v. topology spaces?
Is it just a matter of the geometric component of topology?
Intuitively, I see that when applying a function to these spaces, in the case of topology we may want to relate somehow subsets by proximity; and I can see how in assigning a probability measure in [0,1] to different events, the result will be similar the more related (closer?) these events are…
Post-mortem notes:
- In a $\sigma$-algebra any countable union of elements in $\mathcal{F}$ have to be contained in $\mathcal{F}$. On the other hand for a topological space it is any arbritrary union (finite or infinite) of the elments of the topology $\tau$ has to be contained in $\tau$. $\sim \text{Comment 1 below.}$
- By De Morgan's law and closure under complements of probability spaces, $\mathcal{F}$ is similary closed under countable intersections. Conversely, in topology it is the intersection of finite elements of $\tau$ that remains contained in $\tau$.
Best Answer
No, topologies and measurable spaces are not the same notions ; some topologies on a set are not measurable spaces and some measurable spaces on a set are not topologies.
A topology $T$ usually generates a measurable space $M$ called the Borel $\sigma$-algebra, which is the smallest $\sigma$-algebra containing all open sets, i.e. containing the topology. We have $T=M$ if and only if $T$ is a measurable space, i.e. if and only if $T$ is closed under countable intersections ; this can happen but there are lots of counter-examples (for instance, topological manifolds).
A measurable space $M$ is a topology if and only if it is closed under arbitrary unions, not just countable ones. For instance, if we take an uncountable set $X$ and let $Y \subseteq X$ satisfy $Y \in M$ if and only if $Y$ or $X \backslash Y$ is countable, this defines a $\sigma$-algebra $M$ on $X$. It is not a topology because $X$ admits a subset $Y$ which is uncountable with uncountable complement and we can write $Y = \bigcup_{y \in Y} \{y\}$ (a union of subsets in $M$ which does not belong to $M$).
Hope that helps,