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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.

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  • $\begingroup$ If two laws corresponded to some given function $F_X$, they would coincide on the pi-system $\pi(\mathbb R)$, hence they would in fact be equal, that is, they would coincide on the sigma-algebra $\mathcal B$ generated by this pi-system, end of story. $\endgroup$ – Did Oct 17 '16 at 11:21
  • $\begingroup$ You might as well learn at this point that two measures associated with a measurable space that agree on any collection of subsets (not necessarily a $\pi$ system) that generates the underlying sigma-algebra are equal. $\endgroup$ – Calculon Oct 17 '16 at 11:32
  • $\begingroup$ So are you saying that if $\mathcal{L}_X = F_X$ on $\mathbb{R}$ then $\mathcal{L}_X = F_X$ on $\pi(\mathbb{R})$ and so $\mathcal{L}_X = F_X$ on $\mathcal B$? I'm still a beginner regarding this more abstract theoretical content, sorry. $\endgroup$ – NecroJ Oct 17 '16 at 11:40
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    $\begingroup$ $\mathcal{L}_X$ is a function of the Borel sets. The whole point of this paragraph is that it is sufficient to consider Borel sets of the form $(-\infty,c]$. The reason is as follows. There cannot be two random variables with the same distribution functions, say $F_X = F_Y$, but with different laws, $\mathcal{L}_X \neq \mathcal{L}_Y$. This is the uniqueness lemma. Since the argument of $F_X$ is $c$, it is a lot easier to deal with than $\mathcal{L}_X$. $\endgroup$ – Calculon Oct 17 '16 at 12:33
  • $\begingroup$ One more question (although I basically understand this now): if we only care about $F_X$ on those specific Borel sets, why does it even matter what $\mathcal{L}_X$ does on the Borel sigma algebra? Does it really matter that $\mathcal{L}_X$ doesn't equal $\mathcal{L}_Y$? $\endgroup$ – NecroJ Oct 17 '16 at 14:26
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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

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

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

  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

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

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