Muhammad Haris Rao
Let $X$ be a set, and let $\mathcal{F} \subseteq \mathcal{P}(X)$. Recall the following definitions:
We have the following equivalences:
Lemma: Given a set $X$ and $\mathcal{F} \subseteq \mathcal{P}(X)$, $\mathcal{F}$ is a $\sigma$-algebra if and only if it is both a $\pi$-system and a $\lambda$-system.
Proof. The fact that all $\sigma$-algebras are both $\pi$-systems and $\lambda$-systems is obvious. We only prove here the other implication. So suppose that $\mathcal{F}$ is a $\pi$-system and a $\lambda$-system. We need to prove that $\mathcal{F}$ is closed under countable unions to obtain the desired result. So let $\{ A_n \}_{n \ge 1} \subseteq \mathcal{F}$, and let \begin{align*} B_n = \bigcup_{k = 1}^n A_k = X \big\backslash \bigcap_{k = 1}^n \left( X \backslash A_k \right) \end{align*} We have $B_n \in \mathcal{F}$ because $\mathcal{F}$ is closed under set difference and finite intersections. Then as before, \begin{align*} \bigcup_{n \ge 1} A_n = \bigcup_{n \ge 1} B_n \in \mathcal{F} \end{align*} so $\mathcal{F}$ is a $\sigma$-algebra. $\blacksquare$
Theorem: Let $\Pi \subseteq \mathcal{P}(X)$ be a $\pi$-system, and $\lambda(\Pi)$ the $\lambda$-system generated by $\Pi$. Then $\lambda(\Pi)$ is a $\sigma$-algebra.
Proof. To prove that $\lambda(\Pi)$ is a $\sigma$-algebra, it suffices to show that it is a $\pi$-system since it is already a $\lambda$-system. For $C \in \lambda(\Pi)$, define \begin{align*} \Lambda_C &= \left\{ A \in \lambda(\Pi) \mid A \cap C \in \lambda(\Pi) \right\} \end{align*} Let $C \in \lambda(\Pi)$ be arbitrary. We will prove that $\Lambda_C$ is a $\lambda$-system. Suppose $A, B \in \Lambda_C$ with $A \subseteq B$. Then, \begin{align*} C \cap \left( A \backslash B \right) = \left( C \cap A \right) \backslash \left( C \cap B \right) \end{align*} Because $C \cap A, C \cap B \in \lambda(\Pi)$ by assumption, their set difference as above is also in the $\lambda$-system $\lambda(\Pi)$. Thus, $A \backslash B \in \Lambda_C$ so that $\Lambda_C$ is closed under set differences. Now suppose that $\{ A_n \}_{n \ge 1} \subseteq \Lambda_C$ is an increasing sequence. Then, \begin{align*} C \cap \left( \bigcup_{n \ge 1} A_n \right) = \bigcup_{n \ge 1} \left( C \cap A_n \right) \end{align*} Since $C \cap A_n \in \lambda(\Pi)$, it follows that the above set is also in $\lambda(\Pi)$ since $\lambda$-systems are closed under upper limits of increasing sequences. Thus, \begin{align*} \bigcup_{n \ge 1} A_n \in \Lambda_C \end{align*} This shows that $\Lambda_C$ is a $\lambda$-system. If $C \in \Pi$, then it is easily seen that $\Pi \subseteq \Lambda_C$, so it can only be that $\Lambda_C = \lambda(\Pi)$. Now define \begin{align*} \Lambda^* &= \left\{ C \in \lambda(\Pi) \mid \Lambda_C = \lambda(\Pi) \right\} \end{align*} Evidently, $\Pi \subseteq \Lambda^*$. Now suppose that $A, B \in \Lambda^*$ with $A \subseteq B$. If $C \in \lambda(\Pi)$, then by this assumption we have $A \cap C, B \cap C \in \lambda(\Pi)$. By the definition of a $\lambda$-system, \begin{align*} (A \backslash B) \cap C = \left( A \cap C \right) \backslash \left( B \cap C \right) \in \lambda(\Pi) \end{align*} This proves that $A \backslash B \in \Lambda^*$. Moreover, if $\{ A_n \}_{n \ge 1} \subseteq \Lambda^*$ is an increasing sequence, and $C \in \lambda(\Pi)$ is arbitrary, we have \begin{align*} \left( \bigcup_{n \ge 1} A_n \right) \cap C = \bigcup_{n \ge 1} \left( A_n \cap C \right) \in \lambda(\Pi) \end{align*} This shows that the upper limit of $\{ A_n \}_{n \ge 1}$ is in $\Lambda^*$ as well. Hence, $\Lambda^*$ is a $\lambda$-system. Since it contains $\Pi$ and is contained in $\lambda(\Pi)$, it follows that $\Lambda^* = \lambda(\Pi)$. Then for any $A, B \in \lambda(\Pi)$, we have $A \in \Lambda^*$ which means $\Lambda_A = \lambda(\Pi)$ and so $B \in \Lambda_A$. Recalling the definition of $\Lambda_A$, it follows that $A \cap B \in \lambda(\Pi)$. Since the choice of $A, B \in \lambda(\Pi)$ was arbitrary, it follows that $\lambda(\Pi)$ is a $\pi$-system. Since it is a $\lambda$-system as well, it follows from the previous proposition that $\lambda(\Pi)$ is a $\sigma$-algebra which is what we wanted to show. $\blacksquare$
The consequence of this above fact is that the $\sigma$-algebra and $\lambda$-system generated by a $\pi$-system are the same thing. We have the following corollary for measures:
Theorem: Let $\Pi \subseteq \mathcal{P}(X)$ be a $\pi$-system on a set $X$, and $\mu, \nu : \sigma(\Pi) \longrightarrow [0, \infty]$ measures giving two measure spaces $(X, \sigma(\Pi), \mu)$, $(X, \sigma(\Pi), \nu)$. Suppose further that $\mu(A) = \nu(A)$ for all $A \in \Pi$ and $X$ can be written as a countable union of sets from $\Pi$ each of finite measure. Then $\mu = \nu$.
Proof. Let $\{ C_n \}_{n \ge 1} \subseteq \Pi$ be a disjoint sequence from $\Pi$ such that \begin{align*} X = \bigcup_{n \ge 1} C_n \end{align*} and $\mu(C_n), \nu(C_n) < \infty$ for all $n \ge 1$. Define the set of measurable sets on which $\mu, \nu$ agree to be \begin{align*} \Sigma = \left\{ A \in \sigma(\Pi) \mid \mu(A) = \nu(A) \right\} \end{align*} Evidently, $\Pi \subseteq \Sigma$. We will show that $\Sigma$ is a $\lambda$-system. If $\{ A_n \}_{n \ge 1} \subseteq \Sigma$ is an increasing sequence, then \begin{align*} \mu\left( \bigcup_{n \ge 1} A_n \right) = \lim_{n \to \infty} \mu \left( A_n \right) = \lim_{n \to \infty} \nu \left( A_n \right) = \nu \left( \bigcup_{n \ge 1} A_n \right) \end{align*} so the upper limit of $\{ A_n \}_{n \ge 1}$ is also in $\Sigma$. Thus, $\Sigma$ is closed under upward limits. Now to prove $\Sigma$ is also closed under set differences. For this, we split into two cases.
Suppose first that $\mu(X) < \infty$. Then if $A, B \in \Sigma$ with $B \subseteq A$, \begin{align*} \mu\left( A \backslash B \right) = \mu(A) - \mu(B) = \nu(A) - \nu(B) = \nu \left( A \backslash B \right) \end{align*} so that $A \backslash B \in \Sigma$.
Now to drop the assumption that $\mu(X) < \infty$. If $A, B \in \Sigma$ with $B \subseteq A$, \begin{align*} \mu\left( A \backslash B \right) &= \mu\left( \left( A \backslash B \right) \cap \bigcup_{n \ge 1} C_n \right) \\ &= \mu\left( \bigcup_{n \ge 1} \left( \left( A \backslash B \right) \cap C_n \right) \right) \\ &= \sum_{n \ge 1} \mu \left( \left( A \cap C_n \right) \backslash \left( B \cap C_n \right) \right) \\ &= \sum_{n \ge 1} \left( \mu \left( A \cap C_n \right) - \mu \left( B \cap C_n \right) \right) \\ &= \sum_{n \ge 1} \left( \nu \left( A \cap C_n \right) - \nu \left( B \cap C_n \right) \right) \\ &= \nu\left( A \backslash B \right) \end{align*} So $\Sigma$ is still closed under set differences.
Thus, we have shown that $\Sigma$ is a $\lambda$-system containing $\Pi$. Then $\lambda(\Pi) \subseteq \Sigma$. By the $\pi$-$\lambda$ theorem, \begin{align*} \sigma(\Pi) = \lambda(\Pi) \subseteq \Sigma \subseteq \sigma(\Pi) \end{align*} so really $\Sigma = \sigma(\Pi)$. So $\mu, \nu$ agree on all of $\sigma(\Pi)$ as was claimed. $\blacksquare$
In the language of probability and random variables, this is
Corollary: Let $X, Y$ be real valued random variables. Then $F_X = F_Y$ if and only if $X \overset{d}{=} Y$.