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  "body": "\n## Proof\n\nWe write $e^{itS}$ as a product and argue by induction on the number of factors. We split each complex factor into its real and imaginary parts.\n\n1. Let $X_1, \\dots, X_n$ be [[independent|0XCE070V8B]] real random variables on $(\\Omega, \\mathcal{F}, P)$, and let $S = X_1 + \\dots + X_n$. Fix $t \\in \\R$.\n2. For each $j$, let $U_j = e^{itX_j}$. By [[Euler's formula|04R3CY41VJ]], $U_j = \\cos(tX_j) + i \\sin(tX_j)$, so its real and imaginary parts are bounded by $1$.\n3. For a complex variable $Z = A + iB$ with bounded real variables $A$ and $B$, write $E[Z] = E[A] + i E[B]$. By the definition of the [[characteristic function|0MYDNRT7PP]], $\\varphi_{X_j}(t) = E[U_j]$ and $\\varphi_S(t) = E[e^{itS}]$.\n4. By the [[product of exponentials|0WM7DJCM6G]], $e^{itS} = U_1 U_2 \\cdots U_n$.\n5. For $2 \\le m \\le n$, let $\\mathcal{P}_m$ be the family of events $\\bigcap_{j \u003c m} \\set{X_j \\in B_j}$ with Borel sets $B_j \\subseteq \\R$. Let $\\mathcal{G}_m$ be the σ-algebra generated by $\\mathcal{P}_m$.\n6. The intersection of two such events is $\\bigcap_{j \u003c m} \\set{X_j \\in B_j \\cap B_j'}$, again in $\\mathcal{P}_m$. So $\\mathcal{P}_m$ is a [[π-system|4W9NM803BB]].\n7. Fix a Borel set $B$ and let $C = \\set{X_m \\in B}$. For $A = \\bigcap_{j \u003c m} \\set{X_j \\in B_j}$ in $\\mathcal{P}_m$, the definition of independence with $J = \\set{1, \\dots, m}$ gives $P(A \\cap C) = P(X_m \\in B) \\prod_{j \u003c m} P(X_j \\in B_j)$.\n8. The same definition with $J = \\set{1, \\dots, m - 1}$ gives $\\prod_{j \u003c m} P(X_j \\in B_j) = P(A)$. So $P(A \\cap C) = P(A) P(C)$ for each $A \\in \\mathcal{P}_m$.\n9. Let $\\mathcal{D}$ be the family of events $A$ with $P(A \\cap C) = P(A) P(C)$. It contains $\\Omega$.\n10. If $A \\subseteq A'$ are in $\\mathcal{D}$, then $P((A' \\setminus A) \\cap C) = P(A' \\cap C) - P(A \\cap C) = (P(A') - P(A)) P(C)$. This is $P(A' \\setminus A) P(C)$, so $A' \\setminus A \\in \\mathcal{D}$.\n11. If $A_1 \\subseteq A_2 \\subseteq \\cdots$ are in $\\mathcal{D}$ with union $A$, then $A_k \\cap C$ increases to $A \\cap C$. By [[continuity of probability|0X8HBKCSYB]], $P(A \\cap C) = \\lim_k P(A_k) P(C) = P(A) P(C)$.\n12. By steps 9 to 11, $\\mathcal{D}$ is a [[Dynkin system|69460Z6DMW]] that contains $\\mathcal{P}_m$. By [[Dynkin's π-λ theorem|BBHM1A36M7]], $\\mathcal{G}_m \\subseteq \\mathcal{D}$.\n13. Claim: if $Y$ is a bounded real $\\mathcal{G}_m$-[[measurable|YYNCJSR0PJ]] variable and $g$ is a bounded continuous real function, then $E[Y g(X_m)] = E[Y] \\, E[g(X_m)]$.\n14. Proof of the claim. Each event $\\set{Y \\in D}$ with $D$ Borel lies in $\\mathcal{G}_m$. Each event $\\set{g(X_m) \\in D'}$ equals $\\set{X_m \\in g^{-1}(D')}$, and $g^{-1}(D')$ is Borel.\n15. By step 12, $Y$ and $g(X_m)$ are [[independent|0E9502NT92]]. Both are bounded, hence integrable. The [[product rule for independent variables|05AW1G0WE1]] gives the claim.\n16. Let $V_m = U_1 \\cdots U_{m-1}$, and write $V_m = A_m + i B_m$ with $A_m$, $B_m$ real. Since $\\abs{U_j} = 1$, we have $\\abs{V_m} = 1$, so $A_m$ and $B_m$ are bounded by $1$.\n17. Expanding the product, $A_m$ and $B_m$ are sums of products of the variables $\\cos(tX_j)$ and $\\sin(tX_j)$ with $j \u003c m$. Fix $j \u003c m$ and a Borel set $D$. The set $B_j = \\set{x \\in \\R : \\cos(tx) \\in D}$ is Borel, since $x \\mapsto \\cos(tx)$ is continuous.\n18. So the event $\\set{\\cos(tX_j) \\in D} = \\set{X_j \\in B_j}$ lies in $\\mathcal{P}_m$, with all other sets equal to $\\R$. The same holds for $\\sin$, so these variables are $\\mathcal{G}_m$-measurable.\n19. Sums and products of measurable functions are measurable. So $A_m$ and $B_m$ are $\\mathcal{G}_m$-measurable.\n20. Let $c = \\cos(tX_m)$ and $s = \\sin(tX_m)$. Then $V_m U_m = (A_m c - B_m s) + i (A_m s + B_m c)$.\n21. By [[linearity of expectation|0P7PTC8DDE]] and the claim of step 13, $E[V_m U_m] = (E[A_m] E[c] - E[B_m] E[s]) + i (E[A_m] E[s] + E[B_m] E[c])$.\n22. The right side equals $(E[A_m] + i E[B_m]) (E[c] + i E[s]) = E[V_m] \\, E[U_m]$.\n23. Since $V_m U_m = V_{m+1}$, with $V_{n+1} = U_1 \\cdots U_n$, step 22 gives $E[V_{m+1}] = E[V_m] \\, E[U_m]$ for $2 \\le m \\le n$.\n24. By induction on $m$, starting from $V_2 = U_1$, we get $E[V_{m+1}] = E[U_1] \\cdots E[U_m]$ for $1 \\le m \\le n$.\n25. For $m = n$, steps 3, 4 and 24 give $\\varphi_S(t) = \\varphi_{X_1}(t) \\, \\varphi_{X_2}(t) \\cdots \\varphi_{X_n}(t)$.\n\n## Idea\n\nThe exponential turns the sum $S$ into the product of the factors $e^{itX_j}$. Independence lets the expectation of a product split into the product of the expectations. The π-λ theorem makes the first $m - 1$ factors, taken together, independent of the next one, and the real and imaginary parts are handled separately.\n",
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