Can Sigma Algebras have a Matroid with independent sets defined as mutually independent?

Question

Let ($\Omega$, $\mathcal{A}$, $\mathcal{p}$) be a probability space and consider the pair ($\mathcal{A}$, $\mathcal{I}$) with $\mathcal{I}$ $\subseteq$ $\mathcal{P}$($\mathcal{A}$) where for all $I$ $\in$ $\mathcal{I}$, all members of $I$ are mutually independent. In other words, for all $J$ $\subseteq$ $I$,

$\mathcal{p}$($\bigcap_{A \in {J}}$$A$) $=$ $\prod_{A \in J}$$\mathcal{p}$($A$)

*To clarify on what I mean by independence, if $I$ is a singleton with its only element $A_0$ $\in$ $I$, then $\bigcap${$A_0$} $=$ $A_0$. Thus all singletons are still "independent," or in $\mathcal{I}$, since a single even can satisfy this property with $\mathcal{p}$($\bigcap${$A_0$}) $=$ $\prod_{A \in I}$$\mathcal{p}$($A$) = $\mathcal{p}$($A_0$). Furthermore, I'm not implying that singletons are independent sets in the sense that every event is independent of itself. In other words, this property does not necessarily hold true for all $A$ $\in$ $\mathcal{A}$:

$p$($A$) $=$ $p$($A$ $\cap$ $A$) $=$ $p$($A$)$p$($A$)

implying that $p$($A$) $=$ $0$ or $p$($A$) $=$ $1$

The main question: Is ($\mathcal{A}$, $\mathcal{I}$) a matroid? I already know this satisfies the first two property requirements for $\mathcal{I}$:

1. $\emptyset$ $\in$ $\mathcal{I}$ since the null set is trivially mutually independent
2. For all $J$ $\subseteq$ $I$ and $I$ $\in$ $\mathcal{I}$, $J$ $\in$ $\mathcal{I}$ since all subsets of a mutually independent set is also mutually independent.

But I can't get a proof satisfying this third property for the collection of independent sets:

3. For all $I$,$J$ $\in$ $\mathcal{I}$ and |$I$| $\gt$ |$J$|, there exists a $B$ $\in$ $I$$\backslash$$J$ such that $J$ $\cup$ {$B$} $\in$ $\mathcal{I}$.

Answer 1

Unless I’m misunderstanding something, the final property does not hold. Take for example $\mathcal{A} = \{(a_1,a_2), \; a_i = 0,1\}$, and $p$ uniform measure. This is a model for two independent fair coin flips with e.g. 1 denoting "Heads" and 0 denoting "Tails". Let $J=\{(1,1)\}$, and $I =[ \{(1,0),(1,1)\} , \{(0,1),(1,1)\} ] $. As such $I$ contains the events "the first flip is heads" and "the second flip is heads". $J$ contains the event "both flips are heads". Evidently $J$, containing exactly one event, is in $\mathcal{I}$, with $|J|=1$, $|I|=2$. But the single event in $J$ is dependent on both events in $I$.

Answer 2

This seems a little complicated, can you apply your idea to $\mathcal{M}=(\Omega,\mathcal{A})$? If $\mathcal{M}$ is indeed a matroid what does this imply for your suggested matroid? Note that matroids are motivated (to the best of my knowledge) from combinatorics and lattices, perhaps there are some cardinality issues for abstract spaces.