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  "body": "\n## Proof\n\nWe show that the characteristic function of $\\bar{X}_n$ tends to that of the constant $\\mu$. Then we use Lévy's continuity theorem.\n\n1. Let $\\varphi(t) = E[e^{itX_1}]$ be the [[characteristic function|0MYDNRT7PP]] of $X_1$. The $X_i$ are [[i.i.d.|0AJN7WHBJZ]], so each $X_i$ has the characteristic function $\\varphi$, because it depends only on the law.\n2. Claim: $\\abs{e^{ix} - 1} \\le \\abs{x}$ for real $x$. By the [[fundamental theorem of calculus|0NHGY0N0JB]], applied to the real and imaginary parts, $e^{ix} - 1 = \\int_0^x i e^{iu} \\, du$. The integrand has modulus $1$.\n3. Claim: $\\abs{e^{ix} - 1 - ix} \\le \\min(x^2/2, \\, 2\\abs{x})$ for real $x$. In the same way, $e^{ix} - 1 - ix = \\int_0^x i (e^{iu} - 1) \\, du$, and by step 2 its modulus is at most $\\abs{\\int_0^x \\abs{u} \\, du} = x^2/2$.\n4. Also $\\abs{e^{ix} - 1 - ix} \\le \\abs{e^{ix} - 1} + \\abs{x} \\le 2\\abs{x}$, by step 2. This proves the claim.\n5. For $s \\ne 0$, let $r(s) = E\\left[\\min\\left(\\abs{s} X_1^2/2, \\, 2\\abs{X_1}\\right)\\right]$. By [[linearity of expectation|0P7PTC8DDE]] and step 3 with $x = sX_1$, $\\abs{\\varphi(s) - 1 - i\\mu s} \\le E\\abs{e^{isX_1} - 1 - isX_1} \\le \\abs{s} \\, r(s)$.\n6. Claim: $r(s) \\to 0$ as $s \\to 0$. Let $s_j \\to 0$. The integrand tends to $0$ at each $\\omega$, since it is at most $\\abs{s_j} X_1^2/2$. It is bounded by $2\\abs{X_1}$, which is integrable.\n7. By [[dominated convergence|BYNV84PF86]], $r(s_j) \\to 0$. Since the sequence $s_j$ was arbitrary, the claim holds.\n8. By step 3 with $x = \\mu s$, $\\abs{e^{i\\mu s} - 1 - i\\mu s} \\le \\mu^2 s^2/2$. With step 5, $\\abs{\\varphi(s) - e^{i\\mu s}} \\le \\abs{s} \\, r(s) + \\mu^2 s^2/2$.\n9. 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 $1$.\n10. Let $S_n = X_1 + \\dots + X_n$, so $\\bar{X}_n = S_n/n$. By the definition, $\\varphi_{\\bar{X}_n}(t) = E[e^{i(t/n) S_n}] = \\varphi_{S_n}(t/n)$.\n11. By the [[characteristic function of a sum of independent random variables|02JGYG0T7M]] and step 1, $\\varphi_{\\bar{X}_n}(t) = \\varphi(t/n)^n$.\n12. Fix $t \\ne 0$. By the [[product of exponentials|0WM7DJCM6G]], $e^{i\\mu t} = (e^{i\\mu t/n})^n$. Also $\\abs{\\varphi(t/n)} \\le 1$ and $\\abs{e^{i\\mu t/n}} = 1$.\n13. Steps 9, 11, 12 and 8 give $\\abs{\\varphi_{\\bar{X}_n}(t) - e^{i\\mu t}} \\le n \\abs{\\varphi(t/n) - e^{i\\mu t/n}} \\le \\abs{t} \\, r(t/n) + \\mu^2 t^2/(2n)$.\n14. By step 7, the right side tends to $0$. So $\\varphi_{\\bar{X}_n}(t) \\to \\psi(t) = e^{i\\mu t}$. This also holds for $t = 0$, where both sides equal $1$.\n15. The function $\\psi$ is continuous at $0$. By [[Lévy's continuity theorem|0FPWF7TTMB]], $\\psi$ is the characteristic function of a random variable $Y$, and $\\bar{X}_n \\to Y$ [[in distribution|0JDAQHWA9N]].\n16. The constant random variable $\\mu$ has the characteristic function $E[e^{it\\mu}] = e^{i\\mu t} = \\psi(t)$. Since [[the characteristic function determines the distribution|0T8BYAFEXX]], $P(Y = \\mu) = 1$.\n17. So the [[cumulative distribution function|0MXMS3AE13]] $F_Y$ is $0$ on $(-\\infty, \\mu)$ and $1$ on $[\\mu, \\infty)$. It is continuous at every $x \\ne \\mu$.\n18. Let $F_n$ be the cumulative distribution function of $\\bar{X}_n$, and fix $\\varepsilon \u003e 0$. By step 17, $F_n(\\mu - \\varepsilon) \\to 0$ and $F_n(\\mu + \\varepsilon) \\to 1$.\n19. The event $\\set{\\abs{\\bar{X}_n - \\mu} \u003e \\varepsilon}$ is contained in $\\set{\\bar{X}_n \\le \\mu - \\varepsilon} \\cup \\set{\\bar{X}_n \u003e \\mu + \\varepsilon}$. So $P(\\abs{\\bar{X}_n - \\mu} \u003e \\varepsilon) \\le F_n(\\mu - \\varepsilon) + 1 - F_n(\\mu + \\varepsilon)$.\n20. By step 18, the right side tends to $0$. Hence $\\bar{X}_n \\to \\mu$ [[in probability|0EFDDXTVD6]].\n\n## Idea\n\nA finite mean gives the first-order expansion $\\varphi(s) = 1 + i\\mu s + o(s)$. Raising $\\varphi(t/n)$ to the power $n$ then gives $e^{i\\mu t}$, the characteristic function of the constant $\\mu$. Convergence in distribution to a constant is the same as convergence in probability.\n",
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