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  "body": "\n## Proof\n\n1. Let $(\\Omega, \\mathcal{F}, P)$ be the probability space. Let $X$ and $Y$ be [[random variables|0DKRR22AEQ]] with values in $(E_1, \\mathcal{E}_1)$ and $(E_2, \\mathcal{E}_2)$.\n2. Let $Z \\colon \\Omega \\to E_1 \\times E_2$ be $Z(\\omega) = (X(\\omega), Y(\\omega))$. Let $\\pi_1$ and $\\pi_2$ be the coordinate projections of $E_1 \\times E_2$.\n3. Let $\\mathcal{G}$ be the family of sets $C \\subseteq E_1 \\times E_2$ whose [[preimage|0FGCFY1AZX]] $Z^{-1}(C)$ lies in $\\mathcal{F}$.\n4. $\\mathcal{G}$ is a [[σ-algebra|04WSW8T3SY]]. Indeed $Z^{-1}(E_1 \\times E_2) = \\Omega$, preimages commute with complements, and the preimage of a union is the union of the preimages.\n5. For $A \\in \\mathcal{E}_1$, $\\pi_1^{-1}(A) = A \\times E_2$ and $Z^{-1}(A \\times E_2) = X^{-1}(A) \\in \\mathcal{F}$. In the same way $Z^{-1}(\\pi_2^{-1}(B)) = Y^{-1}(B) \\in \\mathcal{F}$ for $B \\in \\mathcal{E}_2$.\n6. So $\\mathcal{G}$ contains the generators of the [[product σ-algebra|M8X6VZD91K]] $\\mathcal{E}_1 \\otimes \\mathcal{E}_2$. Hence $\\mathcal{E}_1 \\otimes \\mathcal{E}_2 \\subseteq \\mathcal{G}$.\n7. By step 6 the [[joint law|0QAR0XDZH4]] $\\mu_{(X, Y)}(C) = P(Z^{-1}(C))$ is defined for all $C \\in \\mathcal{E}_1 \\otimes \\mathcal{E}_2$.\n8. $\\mu_{(X, Y)}$ is a [[probability measure|0VGMEBVW1M]]. Its total mass is $P(\\Omega) = 1$. Disjoint sets have disjoint preimages, and preimages respect unions, so countable additivity of $P$ passes to $\\mu_{(X, Y)}$.\n9. The same argument shows that the [[laws|0TST4MHA3A]] $\\mu_X(A) = P(X \\in A)$ and $\\mu_Y(B) = P(Y \\in B)$ are probability measures.\n10. They are the [[marginal laws|0AXP2Z5N9H]] of $\\mu_{(X, Y)}$: by step 5, $\\mu_{(X, Y)}(A \\times E_2) = \\mu_X(A)$ and $\\mu_{(X, Y)}(E_1 \\times B) = \\mu_Y(B)$.\n11. Let $A \\in \\mathcal{E}_1$ and $B \\in \\mathcal{E}_2$. Then $A \\times B = \\pi_1^{-1}(A) \\cap \\pi_2^{-1}(B)$ lies in $\\mathcal{E}_1 \\otimes \\mathcal{E}_2$, and $Z^{-1}(A \\times B) = X^{-1}(A) \\cap Y^{-1}(B)$.\n12. So $\\mu_{(X, Y)}(A \\times B) = P(X \\in A, \\, Y \\in B)$ for all $A \\in \\mathcal{E}_1$ and $B \\in \\mathcal{E}_2$.\n13. The measures $\\mu_X$ and $\\mu_Y$ are finite, so they are [[σ-finite|WEBBT18D4X]]. Hence the [[product measure|9ZP4ER2N65]] $\\mu_X \\otimes \\mu_Y$ exists.\n14. By its definition, $\\mu_X \\otimes \\mu_Y$ is the unique measure on $\\mathcal{E}_1 \\otimes \\mathcal{E}_2$ with $(\\mu_X \\otimes \\mu_Y)(A \\times B) = P(X \\in A) \\, P(Y \\in B)$ for all $A$ and $B$.\n15. Suppose that $X$ and $Y$ are [[independent|0E9502NT92]]. Then step 12 gives $\\mu_{(X, Y)}(A \\times B) = P(X \\in A) \\, P(Y \\in B)$ for all $A \\in \\mathcal{E}_1$ and $B \\in \\mathcal{E}_2$.\n16. So $\\mu_{(X, Y)}$ is a measure on $\\mathcal{E}_1 \\otimes \\mathcal{E}_2$ with the property of step 14. By the uniqueness in step 14, $\\mu_{(X, Y)} = \\mu_X \\otimes \\mu_Y$.\n17. Conversely, suppose that $\\mu_{(X, Y)} = \\mu_X \\otimes \\mu_Y$. Evaluating both sides at $A \\times B$ and using steps 12 and 14 gives $P(X \\in A, \\, Y \\in B) = P(X \\in A) \\, P(Y \\in B)$.\n18. This holds for all $A \\in \\mathcal{E}_1$ and $B \\in \\mathcal{E}_2$, so $X$ and $Y$ are independent.\n\n## Idea\n\nOn a rectangle $A \\times B$, the joint law gives $P(X \\in A, Y \\in B)$ and the product measure gives $P(X \\in A) P(Y \\in B)$. Independence says exactly that these agree. A measure on the product σ-algebra is determined by its values on rectangles, so agreement on rectangles is the same as equality of the measures.\n",
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