The weak law of large numbers identifies the deterministic first-order scale of a sum. For independent observations, random fluctuations are smaller than \(n\), so division by \(n\) exposes the mean. The finite-variance proof is a one-line concentration argument; the integrable case shows why truncation is one of the central techniques of probability.
Sample averages and empirical frequencies
Definition. For random variables \(X_1,\ldots,X_n\), put \[ S_n=\sum_{i=1}^nX_i,\qquad \overline X_n=\frac{S_n}{n}. \] A weak law of large numbers with center \(a\in\mathbb R\) is the assertion \(\overline X_n\to a\) in probability; equivalently, for every \(\varepsilon>0\), \[ P(|S_n-na|>\varepsilon n)\longrightarrow0. \]
Taking \(X_i=1_{\{Z_i\in A\}}\) turns an average into the empirical frequency \(P_n(A)=n^{-1}\sum_i1_{\{Z_i\in A\}}\). Thus even the scalar weak law already says that repeated relative frequencies estimate probabilities.
Example (Bernoulli frequencies). If \(X_i\) are independent Bernoulli\((p)\) variables, then \[ P(|\overline X_n-p|>\varepsilon) \le \frac{p(1-p)}{n\varepsilon^2}. \] The estimate gives both convergence and a finite-sample error bound. Hoeffding's inequality later improves the polynomial bound to \(2e^{-2n\varepsilon^2}\), but no exponential inequality is needed for the law itself.
The variance criterion
Proposition. Let \(X_1,X_2,\ldots\) have finite means, and set \(S_n=\sum_{i\le n}X_i\). If \[ \frac{\mathbb ES_n}{n}\longrightarrow a, \qquad \frac{\operatorname{Var}(S_n)}{n^2}\longrightarrow0, \] then \(S_n/n\to a\) in probability.
Proof. For fixed \(\varepsilon>0\), eventually \(|\mathbb ES_n/n-a|<\varepsilon/2\). Chebyshev's inequality then gives \[ P\!\left(\left|\frac{S_n}{n}-a\right|>\varepsilon\right) \le P\!\left(|S_n-\mathbb ES_n|>\frac{\varepsilon n}{2}\right) \le \frac{4\operatorname{Var}(S_n)}{\varepsilon^2n^2}\longrightarrow0. \]
Theorem (Chebyshev weak law). If \(X_i\) are pairwise independent and identically distributed with \(\mathbb EX_1=\mu\) and \(\operatorname{Var}(X_1)=\sigma^2<\infty\), then \[ \overline X_n\longrightarrow\mu\qquad\text{in probability}. \]
Proof. Pairwise independence is enough because it makes every covariance \(\operatorname{Cov}(X_i,X_j)\), \(i\ne j\), vanish. Consequently, \[ \mathbb ES_n=n\mu,\qquad \operatorname{Var}(S_n)=\sum_{i=1}^n\operatorname{Var}(X_i)=n\sigma^2. \] The preceding proposition applies since \(\operatorname{Var}(S_n)/n^2=\sigma^2/n\to0\).
Independence cannot simply be omitted. If \(X_i=Y\) for every \(i\), where \(Y\) is nondegenerate and integrable, then every variable has the same mean and variance but \(\overline X_n=Y\) for all \(n\). What matters is not the marginal distribution alone, but suppression of accumulated covariance.
Khinchin's integrable weak law
Finite variance is convenient but not necessary. The decisive estimate is obtained by cutting observations at the same order as the sum.
Lemma. If \(\mathbb E|X|<\infty\), then \[ tP(|X|>t)\longrightarrow0, \qquad \frac1t\mathbb E\!\left[X^2 1_{\{|X|\le t\}}\right]\longrightarrow0. \]
Proof. On \(\{|X|>t\}\), \(t\le |X|\), hence \[ 0\le tP(|X|>t)\le \mathbb E[|X|1_{\{|X|>t\}}]\to0 \] by dominated convergence. For the second assertion, the integrand \[ \frac{X^2}{t}1_{\{|X|\le t\}} \] converges pointwise to zero and is bounded by \(|X|\). Dominated convergence applies again.
Theorem (Khinchin weak law). If \(X_1,X_2,\ldots\) are i.i.d. and \(\mathbb E|X_1|<\infty\), then \[ \frac1n\sum_{i=1}^nX_i\longrightarrow\mathbb EX_1 \qquad\text{in probability}. \]
Proof. For each row \(n\), truncate all observations at level \(n\): \[ Y_{n,i}=X_i1_{\{|X_i|\le n\}},\qquad 1\le i\le n. \] The union bound and the first part of the lemma yield \[ P\bigl((X_1,\ldots,X_n)\ne(Y_{n,1},\ldots,Y_{n,n})\bigr) \le nP(|X_1|>n)\to0. \tag{1} \] Since the \(Y_{n,i}\) are independent within each row, \[ \operatorname{Var}\!\left(\frac1n\sum_{i=1}^nY_{n,i}\right) =\frac{\operatorname{Var}(Y_{n,1})}{n} \le\frac{\mathbb E[X_1^2 1_{\{|X_1|\le n\}}]}{n}\to0. \tag{2} \] Chebyshev's inequality therefore gives \[ \frac1n\sum_{i=1}^n\bigl(Y_{n,i}-\mathbb EY_{n,i}\bigr)\to0 \quad\text{in probability}. \] Moreover, \(\mathbb EY_{n,1}\to\mathbb EX_1\) by dominated convergence. Combining this convergence with (1) and (2) proves the result.
What the weak law does and does not control
Remark. Convergence in probability controls each sufficiently late average separately: \[ P(|\overline X_n-\mu|>\varepsilon)\to0. \] It does not by itself control the random event that deviations occur infinitely often. That stronger, pathwise question requires summable error probabilities or maximal inequalities and leads to the strong law.
Example (finite mean without finite variance). Suppose \(P(X>x)=x^{-\alpha}\) for \(x\ge1\), with \(1<\alpha<2\). Then \(\mathbb EX<\infty\) but \(\mathbb EX^2=\infty\). The Chebyshev proof is unavailable, while Khinchin's truncation proof still gives \(\overline X_n\to\mathbb EX\) in probability. The fluctuations, however, are generally of order \(n^{1/\alpha}\), not \(\sqrt n\), and their limits belong to stable-law theory.