Suppose we have a set $S$, $I$ is a $\pi$-system on $S$ if $I$ is a set of subsets of $S$ satisfying:
\(\begin{align} I \neq \emptyset \\ I_1, I_2 \in I \implies I_1 \cap I_2 \in I \end{align}\) or in other words, the subsets of $S$ in $I$ are closed under intersection.
This simple system is visually intuitive; we can often work with just an underlying $\pi$-system of a $\sigma$-algebra to prove certain properties about the measure space. I know that’s really vague, so let’s formalize that a bit.
A $\lambda$-system (also called Dynkin system) $D$ is a set of subsets of $S$ satisfying:
\[\begin{align} S \in D \\ A, B \in D, A \subset B \implies B-A \in D \\ (A_n)_n \in D, \bigcup_n A_n = A, A_i \cap A_j = \emptyset \forall i \neq j \implies A \in D \end{align}\]This looks really similar to a $\sigma$-algebra, but isn’t exactly it. Here, we can only make countable union statements on disjoint sets of the space which is a big difference. We can also swap the third statement of a $\lambda$-system with the equivalent statement that:
\[\begin{align} G_n \in D \\ \lim_{n \to \infty} G_n = G \in D \end{align}\]This is true by monotone convergence theorem.
A $\pi$-system $I$ of some set $S$ extended to a $\lambda$-system is denoted $\lambda(I)$ (or $d(I)$ if we use the Dynkin notation). This $\lambda(I)$ is the smallest $\lambda$-system containing $I$. A set $\Sigma$ is both a $\pi$-system and a $\lambda$-system iff it’s a $\sigma$-algebra.
If a set $\Sigma$ is a $\sigma$-algebra, then we have that it’s closed under countable intersections via DeMorgan’s law:
\[\begin{align} \bigcup_n A_n = A \in \Sigma \\ A^c \in \Sigma \\ \implies (\bigcup_n A_n)^c = \bigcap_n A_n^c \in \Sigma \end{align}\]This is a stronger statement than what $\pi$-system states, so it should be one as well. This set should also be a $\lambda$-system because you can just set each $A_n$ in the countable union to be disjoint.
In the other direction, if a set $\Sigma$ is both a $\pi$-system and a $\lambda$-system then we’ll show it’s also a $\sigma$-algebra. Specifically, we’d like to show that countable unions of not necessarily disjoint sets $(A_n)_n$ is in the set (since all the other properties are basically inherited from $\lambda$-systems). Note that:
\[A \cup B = S - (A^c \cap B^c) \in \Sigma\]The complement, set subtraction is inherited from $\lambda$-system (and since $S$ is the whole set, that satisfies the subset requirement). The intersection is inherited from $\pi$-system. This allows us to union two sets even if they’re not disjoint! To finish, we construct a set $G_n := \bigcup_{i \leq n} A_i$ and its limit $G \in \Sigma$ because it’s a $\lambda$-system.
Surprisingly, $\lambda$-systems generated from a $\pi$-system are $\pi$-systems themselves. This result, paired with our previous result, gives us that the generated $\lambda$-system contains a $\sigma$-algebra.
Suppose we have the set:
\[D = \{B \in \lambda(I) : B \cap C \in \lambda (I), \forall C \in \lambda(I)\}\]Is $D$ a $\pi$-system? It should be, since we have by definition that this set is closed under intersection. This set should also be nonempty since $S \cap C = C, \forall C \in \lambda(I)$ so the whole set $S$ must belong in $D$.
Is $D$ a $\lambda$-system? It should be as well. We already know from the above result that $S \in \lambda(I)$. For two sets $B \subset A \in D$, we have to show $A - B \in D$. For all $C \in \lambda(I)$, $(A - B) \cap C$ must be in $D$ for $A - B \in D$. Thankfully,
\[\begin{align} (A - B) \cap C = (A \cap C) - (B \cap C) \quad \text{(distributivity)}\\ A \cap C \in \lambda(I), B \cap C \in \lambda(I) \quad \text{(by definition)}\\ A \cap C \subset B \cap C \quad \text{(intersection preserves subset)}\\ \implies (A \cap C) - (B \cap C) \in \lambda(I) \\ \implies (A-B) \cap C \in D \end{align}\]Finally, for disjoint countable unions of $A_n \in D$, each $A_n \cap C \in \lambda(I) \forall i, n$, so by definition of a $\lambda$-system we have $(\bigcup_n A_n) \cap C \in \lambda(I)$. Now recall $\lambda(I)$ is defined as the smallest $\lambda$-system containing $I$. Here, $D$ is a subset of $\lambda(I)$ but it also is a $\lambda$-system containing $I$, so $D = \lambda(I)$! This means any $\pi$-system extended to a $\lambda$-system is also a $\pi$-system (since $D$ is that $\lambda$-system), and thus a $\sigma$-algebra is always contained within this extension.
Why did we mention all this stuff about $\pi$ and $\lambda$-systems? Because two measures that agree on the same $\pi$-system will be the same measure on the $\sigma$-algebra generated by that $\pi$-system!
To be more specific, suppose we have a $\pi$-system $I$ on a set $S$ and we have two measures $\mu_1, \mu_2$ on $\sigma(I)$, the generated $\sigma$-algebra of the $\pi$-system. If $\mu_1(S) = \mu_2(S) < \infty$ and $\mu_1(A) = \mu_2(A), \forall A \in I$, then $\mu_1(A) = \mu_2(A), \forall A \in \sigma(I)$.
To do this, we need to loop in a $\lambda$-system constructed as:
\[D = \{A \in \sigma(I): \mu_1(A) = \mu_2(A)\}\]Let’s show that it is indeed a $\lambda$-system. Obviously, $S \in \sigma(I)$ for any $I$, and we have that the measures are equal on the whole set. To show that $B - A \in D, \forall A \subset B \in D$, note that
\[\mu_1(B - A) = \mu_1(B) - \mu_1(A) = \mu_2(B) - \mu_2(A) = \mu_2(B - A)\]Since measures of disjoint sets are additive. In fact, measures of disjoint sets are countably additive, which also conveniently proves $\bigcup_n A_n \in D$, if $A_n \in D$ and disjoint (hint: use monotone convergence theorem again). Since we just proved that $D$ is a $\lambda$-system, we know it contains a $\sigma$-algebra within it by our previous result. However, $D \subset \sigma(I)$ by construction, yet it contains $\sigma(I)$, which means $D = \sigma(I)$! This means the measures $\mu_1, \mu_2$ will agree on the generated $\sigma$-algebra (which turns out to be $D$).
Constructing a $\pi$-system is often pretty easy. For reals, we can create a $\pi$-system as:
\[\pi(\mathbb{R}) = \{(-\infty, x] : x \in \mathbb{R} \}\]Compare this to $\mathcal{B}$, the borel $\sigma$-algebra generated by the topology on reals, this is much easier to reason with. With the previous result, we now know that if two probability measures agree on $\pi(\mathbb{R})$ they will also agree on the generated $\sigma(\pi(\mathbb{R})) = \sigma(\mathbb{R}) = \mathcal{B}$.