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  "body": "\n## Proof\n\n1. Let $(\\Omega, \\mathcal{F}, P)$ be a [[probability space|0ZYBZ34RT2]]. Let $(B_i)_{i \\in I}$ be the given family, with $I = \\set{1, \\dots, N}$ or $I = \\set{1, 2, \\dots}$.\n2. So the $B_i$ are pairwise disjoint events with $\\bigcup_{i \\in I} B_i = \\Omega$ and $P(B_i) \u003e 0$. They form a [[partition|09CCE86A6D]] of $\\Omega$. Let $A$ be an event.\n3. For each integer $i \\ge 1$ let $E_i = A \\cap B_i$ if $i \\in I$, and $E_i = \\varnothing$ if $i \\notin I$. Each $E_i$ is in $\\mathcal{F}$, because $\\mathcal{F}$ is a [[σ-algebra|04WSW8T3SY]].\n4. The $E_i$ are pairwise disjoint. For $i \\ne j$ in $I$ we have $E_i \\cap E_j \\subseteq B_i \\cap B_j = \\varnothing$, and the empty sets meet nothing.\n5. Their union is $\\bigcup_{i \\in I} (A \\cap B_i) = A \\cap \\bigcup_{i \\in I} B_i = A \\cap \\Omega = A$.\n6. By the [[complement rule|0PQEBTYDZM]], $P(\\varnothing) = 1 - P(\\Omega) = 0$.\n7. By the countable additivity of the [[probability measure|0VGMEBVW1M]] $P$ and step 6, $P(A) = \\sum_{i \\in I} P(A \\cap B_i)$.\n8. For $i \\in I$ we have $P(B_i) \u003e 0$. By the definition of [[conditional probability|0FJB06AD2P]], $P(A \\cap B_i) = P(A \\mid B_i) \\, P(B_i)$.\n9. Substituting step 8 into step 7 gives $P(A) = \\sum_{i} P(A \\mid B_i) \\, P(B_i)$.\n\n## Idea\n\nThe partition cuts $A$ into the disjoint pieces $A \\cap B_i$, so $P(A)$ is the sum of their probabilities. Each piece has probability $P(A \\mid B_i) P(B_i)$ by the definition of conditional probability.\n",
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