On "for all" in if and only if statements in probability theory and stochastic calculus

Question

1 In my friend's Probability Theory long test there was this question:

Let $(\Omega, \mathfrak{F}, P)$ be a probability space on which is defined all sub-$\sigma$-algebras, events and random variables in this problem.

Define the independence of any two sub-$\sigma$-algebras $\mathfrak{F}_1$, $\mathfrak{F}_2$ of $\mathfrak{F}$.

My friend's answer was:

Let $F_1 \in \mathfrak{F}_1$ and $F_2 \in \mathfrak{F}_2$

Definition:

$\mathfrak{F}_1$ and $\mathfrak{F}_2$ are independent iff $P(F_1 \cap F_2) = P(F_1)P(F_2)$.

Apparently he was deducted a point, and there was the following comment:

"The arbitrary choice of $F_1$ and $F_2$ is part of the 2nd clause."

What exactly is the difference? I have a feeling it has something to do with the logic in the beginning steps of the "if" parts of the proofs of the ff propositions:

2 Also in Probability Theory (my friend is a year ahead of me)

Theorem: Let X and Y be random variables in $(\Omega, \mathfrak{F}, P)$. Then $\mathfrak{L}_X = \mathfrak{L}_Y$ iff $E(g(X)) = E(g(Y)) \forall$ Borel functions g such that the expectations are well-defined.

The proof for the "if" part involved choosing a particular function $g$ and then showing that $\mathfrak{L}_X = \mathfrak{L}_Y$. It seems to me that for that particular function $g$, $\mathfrak{L}_X = \mathfrak{L}_Y$, but it does not necessarily hold true for other functions?

3 In Stochastic Calculus:

Lemma (Martingale Transform) An adapted sequence of real integrable random variables $X_0, X_1, X_2, ...$ is a martingale iff $\forall$ bounded predictable sequences $c_1, c_2, ...$, $E(\sum_{k=1}^{n} c_k \Delta x_k)$

The proof for the "if" part involved choosing a particular sequence $c_k$ and then showing that the sequence $X_k$ was a martingale. It seems to me that for that particular sequence $c_k$, $X_k$ is a martingale, but it does not necessarily hold true for other sequences?

Could it be that my thinking for 2 and 3 is actually instead for the ff statements:

Theorem: $\forall$ Borel functions g such that the expectations are well-defined, let X and Y be random variables in $(\Omega, \mathfrak{F}, P)$. Then $\mathfrak{L}_X = \mathfrak{L}_Y$ iff $E(g(X)) = E(g(Y))$.

Lemma (Martingale Transform) $\forall$ bounded predictable sequence $c_1, c_2, ...$, an adapted sequences of real integrable random variables $X_0, X_1, X_2, ...$ is a martingale iff $E(\sum_{k=1}^{n} c_k \Delta x_k)$

?

If so, why exactly?

Answer 1

Your friend's answer in case 1. is wrong because, according to this "definition", the independence of $\mathfrak{F}_1$ and $\mathfrak{F}_2$ would depend on the choice of $F_1 \in \mathfrak{F}_1$ and $F_2 \in \mathfrak{F}_2$. Thus, this "definition" corresponds to some sort of independence of $\mathfrak{F}_1$ and $\mathfrak{F}_2$ related to $(F_1,F_2)$ in $\mathfrak{F}_1\times\mathfrak{F}_2$, not to an intrinsic property of $\mathfrak{F}_1$ and $\mathfrak{F}_2$.

Imagine for concreteness that $P(F_1 \cap F_2) = P(F_1)P(F_2)$ and that $P(G_1 \cap G_2) \ne P(G_1)P(G_2)$, for some carefully chosen $F_1$ and $G_1$ in $\mathfrak{F}_1$ and $F_2$ and $G_2$ in $\mathfrak{F}_2$. Then what? Using your friend's "definition" twice, $P(F_1 \cap F_2) = P(F_1)P(F_2)$ hence $\mathfrak{F}_1$ and $\mathfrak F_2$ are independent, and $P(G_1 \cap G_2) \ne P(G_1)P(G_2)$ hence $\mathfrak{F}_1$ and $\mathfrak F_2$ are not independent?

By the way, since $\varnothing$ belongs to every sigma-algebra, choosing $F_1=\varnothing$ ensures that $P(F_1 \cap F_2) = P(F_1)P(F_2)$. Thus, every sub-sigma-algebras $\mathfrak{F}_1$ and $\mathfrak{F}_2$ would be independent?

Cases 2. and 3. are examples of the same fallacy.