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  "body": "\n## Proof\n\nWe show that the characteristic function of the normalized sum tends to $e^{-t^2/2}$. Then we use Lévy's continuity theorem.\n\n1. Let $Y_i = (X_i - \\mu)/\\sigma$. For a Borel set $B$, $\\set{Y_i \\in B} = \\set{X_i \\in g^{-1}(B)}$ with the continuous map $g(x) = (x - \\mu)/\\sigma$. So the $Y_i$ are [[independent|0XCE070V8B]] and identically distributed, like the [[i.i.d.|0AJN7WHBJZ]] variables $X_i$.\n2. By [[linearity of expectation|0P7PTC8DDE]] and the definition of the [[variance|007QED3E0Y]], $E[Y_1] = 0$ and $E[Y_1^2] = \\sigma^2/\\sigma^2 = 1$.\n3. Let $Z_n = (X_1 + \\dots + X_n - n\\mu)/(\\sigma\\sqrt{n}) = (Y_1 + \\dots + Y_n)/\\sqrt{n}$. Note that $Z_n = \\sqrt{n}(\\bar{X}_n - \\mu)/\\sigma$.\n4. Let $\\varphi$ be the [[characteristic function|0MYDNRT7PP]] of $Y_1$. It is also the characteristic function of each $Y_i$, since it depends only on the law.\n5. By the definition, $\\varphi_{Z_n}(t) = E[e^{i(t/\\sqrt{n})(Y_1 + \\dots + Y_n)}]$. By the [[characteristic function of a sum of independent random variables|02JGYG0T7M]], $\\varphi_{Z_n}(t) = \\varphi(t/\\sqrt{n})^n$.\n6. Claim: $\\abs{e^{ix} - 1 - ix} \\le x^2/2$ for real $x$. By the [[fundamental theorem of calculus|0NHGY0N0JB]], on real and imaginary parts, $e^{ix} - 1 - ix = \\int_0^x i(e^{iu} - 1) \\, du$. Also $\\abs{e^{iu} - 1} = \\abs{\\int_0^u i e^{iv} \\, dv} \\le \\abs{u}$.\n7. So the modulus in step 6 is at most $\\abs{\\int_0^x \\abs{u} \\, du} = x^2/2$, which proves the claim.\n8. Claim: $\\abs{e^{ix} - 1 - ix + x^2/2} \\le \\min(\\abs{x}^3/6, \\, x^2)$ for real $x$. Again $e^{ix} - 1 - ix + x^2/2 = \\int_0^x i(e^{iu} - 1 - iu) \\, du$, so by step 7 the modulus is at most $\\abs{\\int_0^x u^2/2 \\, du} = \\abs{x}^3/6$.\n9. Also, by step 7, the modulus is at most $\\abs{e^{ix} - 1 - ix} + x^2/2 \\le x^2$. This proves the claim.\n10. For real $s$, let $q(s) = E\\left[\\min\\left(\\abs{s} \\abs{Y_1}^3/6, \\, Y_1^2\\right)\\right]$. By step 2, $\\varphi(s) - 1 + s^2/2 = E[e^{isY_1} - 1 - isY_1 + s^2 Y_1^2/2]$.\n11. By step 8 with $x = sY_1$, $\\abs{\\varphi(s) - 1 + s^2/2} \\le E\\left[\\min(\\abs{s}^3 \\abs{Y_1}^3/6, \\, s^2 Y_1^2)\\right] = s^2 q(s)$.\n12. Claim: $q(s) \\to 0$ as $s \\to 0$. Let $s_j \\to 0$. The integrand tends to $0$ at each $\\omega$, since it is at most $\\abs{s_j} \\abs{Y_1}^3/6$. It is bounded by $Y_1^2$, which is integrable.\n13. By [[dominated convergence|BYNV84PF86]], $q(s_j) \\to 0$. Since the sequence $s_j$ was arbitrary, the claim holds.\n14. Claim: $\\abs{e^{-s^2/2} - 1 + s^2/2} \\le s^4/8$. By [[Taylor's theorem with the Lagrange remainder|0H5ZAZCX2T]], for $y = s^2/2 \u003e 0$ there is $\\xi \\in (0, y)$ with $e^{-y} = 1 - y + e^{-\\xi} y^2/2$. Since $0 \u003c e^{-\\xi} \u003c 1$, the error is at most $y^2/2 = s^4/8$.\n15. Steps 11 and 14 give $\\abs{\\varphi(s) - e^{-s^2/2}} \\le s^2 q(s) + s^4/8$.\n16. Claim: for complex $z, w$ with $\\abs{z} \\le 1$ and $\\abs{w} \\le 1$, $\\abs{z^n - w^n} \\le n \\abs{z - w}$. Indeed $z^n - w^n = (z - w) \\sum_{k=0}^{n-1} z^k w^{n-1-k}$, and each of the $n$ terms has modulus at most $1$.\n17. Fix $t \\in \\R$. By the [[product of exponentials|0WM7DJCM6G]], $e^{-t^2/2} = (e^{-t^2/(2n)})^n$. Also $\\abs{\\varphi(t/\\sqrt{n})} \\le 1$ and $0 \u003c e^{-t^2/(2n)} \\le 1$.\n18. By steps 5, 16, 17 and 15 with $s = t/\\sqrt{n}$, $\\abs{\\varphi_{Z_n}(t) - e^{-t^2/2}} \\le n \\abs{\\varphi(t/\\sqrt{n}) - e^{-t^2/(2n)}} \\le t^2 q(t/\\sqrt{n}) + t^4/(8n)$.\n19. By step 13, the right side tends to $0$. So $\\varphi_{Z_n}(t) \\to \\psi(t) = e^{-t^2/2}$ for each $t \\in \\R$.\n20. The function $\\psi$ is continuous at $0$. By [[Lévy's continuity theorem|0FPWF7TTMB]], $\\psi$ is the characteristic function of a random variable $W$, and $Z_n \\to W$ [[in distribution|0JDAQHWA9N]].\n21. By the [[characteristic function of the normal distribution|0NS0TDA4E2]], $\\psi$ is also the characteristic function of $\\mathcal{N}(0, 1)$. Since [[the characteristic function determines the distribution|0T8BYAFEXX]], $W \\sim \\mathcal{N}(0, 1)$, and its cumulative distribution function is $\\Phi$.\n22. The [[standard normal density|07RXBVTVZB]] is bounded by $1/\\sqrt{2\\pi}$. So $\\abs{\\Phi(x) - \\Phi(y)} \\le \\abs{x - y}/\\sqrt{2\\pi}$, and $\\Phi$ is continuous at every point.\n23. By steps 20 to 22, $P(Z_n \\le x) \\to \\Phi(x)$ for each $x \\in \\R$. This is the claim of the theorem.\n\n## Idea\n\nAfter centering and scaling, $Y_1$ has characteristic function $1 - s^2/2 + o(s^2)$, the same to second order as $e^{-s^2/2}$. Raising to the power $n$ at $s = t/\\sqrt{n}$ multiplies the error by $n$, but the error is $o(1/n)$, so the limit is $e^{-t^2/2}$. Lévy's continuity theorem turns this into convergence of the distribution functions.\n",
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