Proof of Dynkin's Pi-Lambda Theorem

Let $\mathcal{L}_0$ be the intersection of all $\lambda$-systems on $X$ containing $\mathcal{P}$; the collection is nonempty (the family of all subsets of $X$ is a $\lambda$-system), and the intersection of $\lambda$-systems is a $\lambda$-system, since each of the three defining properties is preserved under intersections of families. Thus $\mathcal{L}_0$ is the smallest $\lambda$-system containing $\mathcal{P}$, and $\mathcal{L}_0\subseteq\mathcal{L}$. It suffices to show $\sigma(\mathcal{P})\subseteq\mathcal{L}_0$.

Step 1 ($\mathcal{L}_0$ is a $\pi$-system). First let

$\mathcal{L}_1$ contains $\mathcal{P}$ (as $\mathcal{P}$ is a $\pi$-system and $\mathcal{P}\subseteq\mathcal{L}_0$), and $\mathcal{L}_1$ is a $\lambda$-system: $X\cap C=C\in\mathcal{L}_0$; if $A\subseteq B$ both lie in $\mathcal{L}_1$ then $(B\setminus A)\cap C=(B\cap C)\setminus(A\cap C)$ with $A\cap C\subseteq B\cap C$ both in $\mathcal{L}_0$, so the relative complement lies in $\mathcal{L}_0$ by property 2; and for a nondecreasing sequence $(A_m)$ in $\mathcal{L}_1$, $\bigl(\bigcup_m A_m\bigr)\cap C=\bigcup_m(A_m\cap C)$ is a nondecreasing union of members of $\mathcal{L}_0$, hence in $\mathcal{L}_0$ by property 3. By minimality of $\mathcal{L}_0$, $\mathcal{L}_0\subseteq\mathcal{L}_1$.

Next let

By the previous paragraph, $\mathcal{P}\subseteq\mathcal{L}_2$: for $C\in\mathcal{P}$ and $B\in\mathcal{L}_0\subseteq\mathcal{L}_1$ we have $C\cap B=B\cap C\in\mathcal{L}_0$. The same set identities as above show $\mathcal{L}_2$ is a $\lambda$-system, so $\mathcal{L}_0\subseteq\mathcal{L}_2$ by minimality; that is, $\mathcal{L}_0$ is closed under finite intersections, hence a $\pi$-system (it is nonempty, containing $X$).

Step 2 (a family that is both a $\pi$-system and a $\lambda$-system is a $\sigma$-algebra). $X\in\mathcal{L}_0$ by $\lambda$-property 1; complements: $X\setminus A$ is the relative complement of $A\subseteq X$ in $X$, in $\mathcal{L}_0$ by $\lambda$-property 2; finite unions: $A\cup B=X\setminus\bigl((X\setminus A)\cap(X\setminus B)\bigr)$, using complements and the $\pi$-property; countable unions: for a sequence $(A_m)$ in $\mathcal{L}_0$, the finite unions $F_k=A_1\cup\cdots\cup A_k$ lie in $\mathcal{L}_0$ and form a nondecreasing sequence with $\bigcup_k F_k=\bigcup_m A_m$, which lies in $\mathcal{L}_0$ by $\lambda$-property 3. Hence $\mathcal{L}_0$ satisfies the three properties of a $\sigma$-algebra.

Conclusion. $\mathcal{L}_0$ is a $\sigma$-algebra containing $\mathcal{P}$, so the generated $\sigma$-algebra satisfies $\sigma(\mathcal{P})\subseteq\mathcal{L}_0\subseteq\mathcal{L}$. $\blacksquare$