D. Williams 'Probability with Martingales' query I don't understand the logic of the paragraph highlighted in red at all. The 'uniqueness lemma' he speaks of states that if two measures agree on a so-called $\pi$-system then they agree on the sigma algebra generated by that system.

So is he saying that the two measures in question, $\mathcal{L}_X$ and $\mathbb{P}$, agree on $\pi(\mathbb{R})$? If so, I don't understand why. I'm generally really confused by the number of sigma algebras in this page (we have $\mathcal{F}$, $\mathcal{B}$, $\sigma(X)$ - why so many and what is happening?) and what everything has a function for.

The steps might be really simple for most people like Williams but I can't understand how that lemma relates to this? If anyone can provide some insight into understanding this I would be really grateful.
The first part is saying that if

$$(\Omega, \mathscr F, \mathbb P)$$ is a probability space

then

$$(\mathbb R, \mathscr B, \mathcal L)$$

is a probability space because $\mathcal L$ is a probability measure on (\mathbb R, \mathscr B).$

This can be shown by proving


$\mathcal L(\mathbb R) = 1$:


$P(X \in \mathbb R) = P(\Omega) = 1$


$\mathcal L$ is $\sigma$-additive:


$\mathbb P(X \in B)$ for $B \in \mathscr B$ is $\sigma$-additive because $\mathbb P(A)$ for $A \in \mathscr F$ is $\sigma$-additive



What the second part is saying is that

$$\mathcal L_X(B) := P(X \in B)$$

is determined uniquely by

$$F_X(t) := P(X \le t) = P(X \in (-\infty,t])$$

In other words, if we had some formula for $P(X \in (-\infty,t])$ for any $t \in \mathbb R$ (or equivalently $P(X \in B)$ for any $B \in \mathbb \pi(\mathbb R)$) then we would be able to compute $P(X \in B)$ for any $B \in \mathscr B$, the latter of which is more difficult to compute than the former.

Why? Well as with any proof for uniqueness suppose we have two random variables $X$ and $Y$ with the same distribution function. Is it possible that

$$P(X \in B) = P(Y \in B), \forall B \in \pi(\mathbb R)$$

but

$$\exists B \in \mathscr B \ \text{s.t.} \ P(X \in B) \ne P(Y \in B)$$

?

No. By the uniqueness lemma,

since $$F_X(t) = F_Y(t), t \in \mathbb R$$

or equivalently

$$P(X \in (-\infty,t]) = P(Y \in (-\infty,t]), t \in \mathbb R$$

or equivalently

$$P(X \in B) = P(Y \in B), B \in \pi(\mathbb R)$$

we have that

$$P(X \in B) = P(Y \in B), B \in \sigma(\pi(\mathbb R)) = \mathscr B$$
The first part is saying that if

$$(\Omega, \mathscr F, \mathbb P)$$ is a probability space

then

$$(\mathbb R, \mathscr B, \mathcal L)$$

is a probability space because $\mathcal L$ is a probability measure on (\mathbb R, \mathscr B).$

This can be shown by proving


$\mathcal L(\mathbb R) = 1$:


$P(X \in \mathbb R) = P(\Omega) = 1$


$\mathcal L$ is $\sigma$-additive:


$\mathbb P(X \in B)$ for $B \in \mathscr B$ is $\sigma$-additive because $\mathbb P(A)$ for $A \in \mathscr F$ is $\sigma$-additive



What the second part is saying is that

$$\mathcal L_X(B) := P(X \in B)$$

is determined uniquely by

$$F_X(t) := P(X \le t) = P(X \in (-\infty,t])$$

In other words, if we had some formula for $P(X \in (-\infty,t])$ for any $t \in \mathbb R$ (or equivalently $P(X \in B)$ for any $B \in \mathbb \pi(\mathbb R)$) then we would be able to compute $P(X \in B)$ for any $B \in \mathscr B$, the latter of which is more difficult to compute than the former.

Why? Well as with any proof for uniqueness suppose we have two random variables $X$ and $Y$ with the same distribution function. Is it possible that

$$P(X \in B) = P(Y \in B), \forall B \in \pi(\mathbb R)$$

but

$$\exists B \in \mathscr B \ \text{s.t.} \ P(X \in B) \ne P(Y \in B)$$

?

No. By the uniqueness lemma,

since $$F_X(t) = F_Y(t), t \in \mathbb R$$

or equivalently

$$P(X \in (-\infty,t]) = P(Y \in (-\infty,t]), t \in \mathbb R$$

or equivalently

$$P(X \in B) = P(Y \in B), B \in \pi(\mathbb R)$$

we have that

$$P(X \in B) = P(Y \in B), B \in \sigma(\pi(\mathbb R)) = \mathscr B$$