TheoremBase

Proof

Step 0 (reduction to mean zero). Set Ym=Xm−μY_m=X_m-\mu. The sequence (Ym)(Y_m) is independent and identically distributed: translation by −μ-\mu is a continuous, hence Borel measurable, map (generator criterion of Measurable Function and Real-Valued Measurable Function), so the defining events {Ym∈B}\{Y_m\in B\} are events of the form {Xm∈B′}\{X_m\in B'\} for Borel B′B', and both the independence product formula and equality of distributions transfer. Moreover E[Y1]=0\mathbb{E}[Y_1]=0, and Y14Y_1^{4} has finite expectation since (x−μ)4(x-\mu)^{4} expands binomially into powers xkx^{k}, k≤4k\le 4, each with finite expectation by hypothesis and linearity (claim 2 of Linearity and Monotonicity of the Lebesgue Integral). Writing Tn=Y1+⋯+Yn=Sn−nμT_n=Y_1+\cdots+Y_n=S_n-n\mu, it suffices to prove Tn/n→0T_n/n\to 0 almost surely.

Step 1 (fourth moment bound). Expanding Tn4T_n^{4} multinomially and using linearity,

E[Tn4]=∑i,j,k,l=1nE[YiYjYkYl].\mathbb{E}[T_n^{4}]=\sum_{i,j,k,l=1}^{n}\mathbb{E}[Y_iY_jY_kY_l].

For a term whose index multiset contains an index ii appearing exactly once, group the remaining three factors into the single random variable ZZ (a product of powers of the YY's with indices different from ii); the pair (Yi,Z)(Y_i,Z) satisfies the independence hypothesis of Expectation of a Product of Independent Random Variables, since ZZ is obtained from the finitely many YrY_r, r≠ir\ne i, by products and powers, and the events {Z∈B}\{Z\in B\} lie in the collection generated by events {Yr∈Br}\{Y_r\in B_r\}, r≠ir\ne i, for which the product formula with events of YiY_i holds by independence of finite subfamilies; integrability of the factors holds by the moment bounds of Step 0. Hence such terms equal E[Yi] E[Z]=0\mathbb{E}[Y_i]\,\mathbb{E}[Z]=0. The surviving terms are the nn diagonal terms E[Yi4]\mathbb{E}[Y_i^{4}] and the 3n(n−1)3n(n-1) terms E[Yi2Yj2]\mathbb{E}[Y_i^{2}Y_j^{2}] with i≠ji\ne j, each equal to E[Y12]2\mathbb{E}[Y_1^{2}]^{2} by Expectation of a Product of Independent Random Variables and Step 0 of the proof of Weak Law of Large Numbers (identically distributed variables share moments). Therefore, with C=E[Y14]+3 E[Y12]2C=\mathbb{E}[Y_1^{4}]+3\,\mathbb{E}[Y_1^{2}]^{2},

E[Tn4]=n E[Y14]+3n(n−1) E[Y12]2 ≤ C n2.\mathbb{E}[T_n^{4}]=n\,\mathbb{E}[Y_1^{4}]+3n(n-1)\,\mathbb{E}[Y_1^{2}]^{2}\ \le\ C\,n^{2}.

Step 2 (Borel–Cantelli). Fix ε>0\varepsilon>0. By Markov's inequality (Markov's and Chebyshev's Inequalities) applied to the nonnegative random variable Tn4T_n^{4} with threshold n4ε4n^{4}\varepsilon^{4},

P(∣Tn/n∣≥ε)=P(Tn4≥n4ε4) ≤ Cn2n4ε4=Cn2ε4,P\bigl(|T_n/n|\ge\varepsilon\bigr)=P\bigl(T_n^{4}\ge n^{4}\varepsilon^{4}\bigr)\ \le\ \frac{C n^{2}}{n^{4}\varepsilon^{4}}=\frac{C}{n^{2}\varepsilon^{4}},

and ∑nC/(n2ε4)<∞\sum_n C/(n^{2}\varepsilon^{4})<\infty (the partial sums of ∑1/n2\sum 1/n^{2} are bounded, e.g. by 22, via 1/n2≤1/(n(n−1))=1/(n−1)−1/n1/n^{2}\le 1/(n(n-1))=1/(n-1)-1/n for n≥2n\ge 2). By the first Borel–Cantelli lemma, for each j∈Nj\in\mathbb{N} the event Nj=lim sup⁡n{∣Tn/n∣≥1/j}N_j=\limsup_n\{|T_n/n|\ge 1/j\} has P(Nj)=0P(N_j)=0.

Step 3 (conclusion). Let N=⋃jNjN=\bigcup_j N_j; by countable subadditivity of PP (disjointify and use countable additivity with monotonicity, as in claim 1 of the proof of Existence of Lebesgue Measure on the Real Line for measures generally), P(N)=0P(N)=0. For ω∉N\omega\notin N: for every jj there are only finitely many nn with ∣Tn(ω)/n∣≥1/j|T_n(\omega)/n|\ge 1/j, which is precisely the statement that Tn(ω)/n→0T_n(\omega)/n\to 0 in the sense of Limit of a Sequence of Real Numbers. Hence the convergence set contains the complement of NN and has probability 11: Tn/n→0T_n/n\to 0, i.e. Sn/n→μS_n/n\to\mu, almost surely. ■\blacksquare

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