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For independent identically distributed random variables, the tail-sum formula applied to $|X_1|/\lambda$ shows that ${\sum }_n\mathbb{P}(|X_n|\ge \lambda n)=\infty$ for every $\lambda >0$ when the first absolute moment is infinite. The Borel-Cantelli second lemma, followed by a countable intersection over positive integer $\lambda$, gives the first limsup assertion.

If $|S_n|/n$ had finite limsup along such a sample path, it would eventually be bounded by some finite constant $C$. Then $X_n=S_n-S_{n-1}$ would imply $|X_n|/n\le 2C$ eventually, a contradiction. This is a necessary-integrability counterpart to the strong law of large numbers; without a finite first absolute moment, the normalized sums cannot converge to any finite value.

On $\{0,1{\}}^{\mathbb{N}}$, the event of infinitely many heads isIt says that after every index there is another head. For each $n$, the same intersection can start at $m=n$, so $A\in \sigma (X_n,X_{n+1},\dots )$; therefore $A$ is a tail event. This verifies measurability directly using countable unions and intersections.

Assume now that the coordinates are fair independent Bernoulli trials. For fixed $m$, let $B_{m,r}$ be the event of at least one head among $X_m,\dots ,X_{m+r-1}$. Its complement requires $r$ tails, soThe last equality follows from continuity of probability. Now $A={\bigcap }_{m\ge 1}{B}_m$, a countable intersection of almost sure events, and hence $\mathbb{P}(A)=1$. This determines which of the two possibilities in Kolmogorov's zero–one law occurs. Equivalently, the Borel-Cantelli second lemma applies to the independent events $\{X_j=1\}$.

Nevertheless, no finite prefix certifies membership in $A$: every finite record can be extended by all tails or by alternating heads and tails. Under the fair independent model, conditioning on any exact finite prefix leaves $\mathbb{P}(A)=1$, since $A$ is independent of the initial sigma-algebra. The excluded all-tails continuation has probability zero, but remains a possible member of the sample space. Logical determination by a finite record and a conclusion holding almost surely are different statements.

For each positive integer $m$, the events $\{|X_n|\ge mn\}$ are independent and have divergent total probability by part (i). The Borel-Cantelli second lemma makes them occur infinitely often. Taking the countable intersection over $m$ givesIf ${\mathrm{limsup}}_n|S_n|/n$ were finite along any such sample path, then for some finite path-dependent $C$, $|S_n|\le Cn$ eventually. But $X_n=S_n-S_{n-1}$ would give $|X_n|/n\le C+C(n-1)/n\le 2C$ eventually, a contradiction. ThereforeThis is the divergence of normalized iid partial sums without a first absolute moment.

The Borel-Cantelli second lemma says that independent events whose probabilities have divergent sum occur infinitely often almost surely. The events here are independent and have probabilities $\min \{1,\lambda /n\}$. Their sum diverges by the harmonic series, so

For events $E_n$, the Borel-Cantelli first lemma states that ${\sum }_n\mathbb{P}(E_n)<\infty$ implies $\mathbb{P}(E_n\text{ infinitely often})=0$. Indeed,These unions decrease to $\mathrm{limsup}{E}_n$, so continuity from above proves the assertion.

The Borel-Cantelli second lemma states that independent events with ${\sum }_n\mathbb{P}(E_n)=\infty$ satisfy $\mathbb{P}(E_n\text{ infinitely often})=1$. For every fixed $m$, independence and $1-x\le e^{-x}$ giveContinuity from above shows there is probability zero of no event after $m$. A countable union over $m$ gives probability zero of only finitely many occurrences. Independence is needed for this second conclusion, but not for the first.

Let $E_n=\{|X_n-1|=n\}$. These events are independent and ${\sum }_{n\ge 2}\mathbb{P}(E_n)={\sum }_{n\ge 2}1/(n\log n)=\infty$. By the Borel-Cantelli second lemma, they occur infinitely often almost surely.

If $S_n/n$ had any finite sample-path limit, thenBut on $E_{n+1}$, $|X_{n+1}|/n$ is either $1$ or $(n+2)/n$, contradicting this convergence infinitely often. Therefore the averages have no finite almost sure limit. This is a rare-jump counterexample to a strong law with convergent first-moment averages.

Both colour counts tend to infinity almost surely. For example, $X_n\ge B_0/(m+n)$, and independent events $U_{n+1}\le B_0/[2(n+1)]$ occur infinitely often by the Borel-Cantelli second lemma, forcing infinitely many blue additions.

The martingale increment formula gives $\mathbb{E}[(X_{k+1}-X_k{)}^2\mid {\mathcal{F}}_k]\le X_k/(m+k+1{)}^2$. Conditional orthogonality of later increments and $\mathbb{E}[X_k\mid {\mathcal{F}}_n]=X_n$ implyOn $X_{\infty }=0$ the squared difference is $X_n^2$, so its conditional probability is at most $1/[(m+n)X_n]=1/B_n\to 0$. Dominated convergence proves zero probability of that endpoint. The symmetric argument for red excludes $X_{\infty }=1$. No explicit beta law is needed for this proof.

Take independent random variables equal to $1+n$ and $1-n$ with probability $1/(2n\log n)$ each, and otherwise equal to one. Their first absolute deviations are $1/\log n\to 0$, so $X_n\to 1$ in L1 convergence. The same triangle-inequality argument gives convergence in $L^1$ of their arithmetic averages to one. For $p>1$, however, $\mathbb{E}|X_n-1{|}^p=n^{p-1}/\log n\to \infty$.

The exceptional jumps occur infinitely often almost surely by the Borel-Cantelli second lemma, since $\sum 1/(n\log n)=\infty$. If the arithmetic averages $S_n/n$ converged to any finite value along a sample path, then $X_{n+1}/n=S_n/n-[(n-1)/n][S_{n-1}/(n-1)]$ would tend to zero. At each exceptional jump its magnitude is at least one. Thus the averages have no finite almost sure limit. Independence and convergence of first-moment averages do not alone give a strong law of large numbers.