[proofplan] We show that the smallest d-system $\mathcal{D}_0 = d(\mathcal{A})$ containing the $\pi$-system $\mathcal{A}$ is itself a $\sigma$-algebra, from which $\sigma(\mathcal{A}) \subseteq \mathcal{D}_0$ follows by minimality. The key fact is that a collection which is simultaneously a $\pi$-system and a d-system is automatically a $\sigma$-algebra. We bootstrap the intersection-closure property from $\mathcal{A}$ to $\mathcal{D}_0$ in two stages: first we show that for each $B \in \mathcal{A}$, the "good set" collection $\mathcal{G}_B = \{A \in \mathcal{D}_0 : A \cap B \in \mathcal{D}_0\}$ is a d-system containing $\mathcal{A}$; then we repeat the argument with $B$ ranging over $\mathcal{D}_0$. [/proofplan] [step:Observe that $\mathcal{D}_0 \subseteq \sigma(\mathcal{A})$ by minimality] Let $\mathcal{D}_0 = d(\mathcal{A})$ denote the smallest d-system containing $\mathcal{A}$. Since every $\sigma$-algebra is a d-system, and $\sigma(\mathcal{A})$ is a d-system containing $\mathcal{A}$, we have $\mathcal{D}_0 \subseteq \sigma(\mathcal{A})$. It therefore suffices to prove that $\mathcal{D}_0$ is a $\sigma$-algebra, for then $\sigma(\mathcal{A}) \subseteq \mathcal{D}_0$ by minimality of $\sigma(\mathcal{A})$. [/step] [step:Show that a collection which is both a $\pi$-system and a d-system is a $\sigma$-algebra] [claim:Pi System Plus D System Equals Sigma Algebra] A collection $\mathcal{C}$ of subsets of $E$ that is both a $\pi$-system and a d-system is a $\sigma$-algebra. [/claim] [proof] Since $\mathcal{C}$ is a d-system, $E \in \mathcal{C}$ and $\mathcal{C}$ is closed under proper differences and increasing countable unions. Since $\mathcal{C}$ is a $\pi$-system, it is closed under finite intersections. We verify the $\sigma$-algebra axioms. First, $\emptyset \in \mathcal{C}$ since $\mathcal{C}$ is a $\pi$-system. Second, if $A \in \mathcal{C}$, then $A \subseteq E$ and both belong to $\mathcal{C}$, so $A^c = E \setminus A \in \mathcal{C}$ by the d-system property. Third, let $(A_n)_{n=1}^\infty$ be an arbitrary sequence in $\mathcal{C}$. Define $B_n = A_1 \cup \cdots \cup A_n$. Using De Morgan's law, $B_n = (A_1^c \cap \cdots \cap A_n^c)^c$. Since $\mathcal{C}$ is closed under complements and finite intersections, $B_n \in \mathcal{C}$ for each $n$. The sequence $(B_n)$ is increasing and $\bigcup_n A_n = \bigcup_n B_n$, so by the d-system property (closure under increasing unions), $\bigcup_n A_n \in \mathcal{C}$. [/proof] By this claim, it suffices to show that $\mathcal{D}_0$ is a $\pi$-system. [/step] [step:Bootstrap intersection-closure to $\mathcal{D}_0$ by fixing elements of $\mathcal{A}$] For each $B \subseteq E$, define \begin{align*} \mathcal{G}_B = \{ A \in \mathcal{D}_0 : A \cap B \in \mathcal{D}_0 \}. \end{align*} [claim:G Is A D System For Pi System Elements] For every $B \in \mathcal{A}$, the collection $\mathcal{G}_B$ is a d-system containing $\mathcal{A}$. [/claim] [proof] Fix $B \in \mathcal{A}$. Since $\mathcal{A}$ is a $\pi$-system and $\mathcal{A} \subseteq \mathcal{D}_0$, for any $A \in \mathcal{A}$ we have $A \cap B \in \mathcal{A} \subseteq \mathcal{D}_0$, so $A \in \mathcal{G}_B$. Thus $\mathcal{A} \subseteq \mathcal{G}_B$. We verify the d-system axioms for $\mathcal{G}_B$. First, $E \cap B = B \in \mathcal{A} \subseteq \mathcal{D}_0$, so $E \in \mathcal{G}_B$. Second, if $A_1, A_2 \in \mathcal{G}_B$ with $A_1 \subseteq A_2$, then $(A_2 \setminus A_1) \cap B = (A_2 \cap B) \setminus (A_1 \cap B)$. Since $A_1 \cap B \subseteq A_2 \cap B$ and both belong to $\mathcal{D}_0$, the d-system property of $\mathcal{D}_0$ gives $(A_2 \setminus A_1) \cap B \in \mathcal{D}_0$, so $A_2 \setminus A_1 \in \mathcal{G}_B$. Third, if $(A_n)$ is increasing in $\mathcal{G}_B$, then $(A_n \cap B)$ is increasing in $\mathcal{D}_0$, so $(\bigcup_n A_n) \cap B = \bigcup_n (A_n \cap B) \in \mathcal{D}_0$, giving $\bigcup_n A_n \in \mathcal{G}_B$. [/proof] Since $\mathcal{G}_B$ is a d-system containing $\mathcal{A}$ and $\mathcal{D}_0$ is the smallest such, $\mathcal{D}_0 \subseteq \mathcal{G}_B$ for every $B \in \mathcal{A}$. This means: for every $A \in \mathcal{D}_0$ and $B \in \mathcal{A}$, we have $A \cap B \in \mathcal{D}_0$. [/step] [step:Extend intersection-closure to all pairs in $\mathcal{D}_0$] [claim:G Is A D System For All D Zero Elements] For every $B \in \mathcal{D}_0$, the collection $\mathcal{G}_B$ is a d-system containing $\mathcal{A}$. [/claim] [proof] Fix $B \in \mathcal{D}_0$. For any $A \in \mathcal{A} \subseteq \mathcal{D}_0$, the previous step gives $A \cap B \in \mathcal{D}_0$ (since $A \in \mathcal{D}_0$ and $B \in \mathcal{D}_0$, and we showed $\mathcal{D}_0 \subseteq \mathcal{G}_A$ for $A \in \mathcal{A}$, meaning $B \in \mathcal{G}_A$, i.e., $A \cap B \in \mathcal{D}_0$). So $\mathcal{A} \subseteq \mathcal{G}_B$. The verification that $\mathcal{G}_B$ is a d-system proceeds identically to the previous claim: the d-system axioms for $\mathcal{G}_B$ follow from those of $\mathcal{D}_0$ and the distributivity of intersection over unions and differences. [/proof] Again, $\mathcal{D}_0 \subseteq \mathcal{G}_B$ for every $B \in \mathcal{D}_0$. This means: for every $A, B \in \mathcal{D}_0$, $A \cap B \in \mathcal{D}_0$. Hence $\mathcal{D}_0$ is a $\pi$-system. [/step] [step:Conclude that $\sigma(\mathcal{A}) \subseteq \mathcal{D}$ for any d-system $\mathcal{D} \supseteq \mathcal{A}$] By the preceding steps, $\mathcal{D}_0$ is both a $\pi$-system and a d-system, hence a $\sigma$-algebra containing $\mathcal{A}$. By minimality, $\sigma(\mathcal{A}) \subseteq \mathcal{D}_0$. Combined with $\mathcal{D}_0 \subseteq \sigma(\mathcal{A})$ from the first step, we get $\mathcal{D}_0 = \sigma(\mathcal{A})$. Since any d-system $\mathcal{D} \supseteq \mathcal{A}$ satisfies $\mathcal{D}_0 \subseteq \mathcal{D}$, we conclude $\sigma(\mathcal{A}) \subseteq \mathcal{D}$. [/step]