Borel–Cantelli Lemmas

$\limsup_{n \to \infty} E_n = \bigcap_{N \ge 1} \bigcup_{k \ge N} E_k$ is the set of points belonging to infinitely many $E_k$.

Since $E_k = [0, \frac{1}{k^2}]$ is decreasing, $\bigcup_{k \ge N} E_k = E_N = [0, \frac{1}{N^2}]$, and $\bigcap_{N \ge 1} [0, \frac{1}{N^2}] = \{0\}$ (any nonzero point is dropped once $\frac{1}{N^2}$ falls below it).

Hence $\limsup E_n = \{0\}$ — a single point of measure 0.

Now the measure bound.

For each $N$, $\limsup E_n \subseteq \bigcup_{k \ge N} E_k$, so by countable subadditivity $m(\limsup E_n) \le \sum_{k \ge N} m(E_k) = \sum_{k \ge N} \frac{1}{k^2}$.

This holds for every $N$.

First ground the convergence of $\sum_{k \ge 1} \frac{1}{k^2}$ self-containedly: for $k \ge 2$, $\frac{1}{k^2} \le \frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k}$, so the partial sums telescope: $\sum_{k=1}^{N} \frac{1}{k^2} \le 1 + \sum_{k=2}^{N} \left(\frac{1}{k-1} - \frac{1}{k}\right) = 2 - \frac{1}{N} \le 2$.

Hence the series converges (its value is $\frac{\pi^2}{6}$) and the tail sums vanish: for every $\varepsilon > 0$ there is $N$ with $\sum_{k \ge N} \frac{1}{k^2} < \varepsilon$.

Hence $m(\limsup E_n) < \varepsilon$ for every $\varepsilon > 0$, so $m(\limsup E_n) = 0$ — matching the direct computation $\{0\}$. That is the first Borel–Cantelli lemma on this example.

Write $\limsup_{n \to \infty} E_n = \bigcap_{N \ge 1} A_N$ where $A_N = \bigcup_{k \ge N} E_k$ is a decreasing sequence (larger $N$ means fewer terms in the union). By definition, $\limsup E_n \subseteq A_N$ for every $N$.

By monotonicity and countable subadditivity, $m(\limsup E_n) \le m(A_N) \le \sum_{k \ge N} m(E_k)$. The second inequality is just that the measure of a countable union never exceeds the sum of the measures — no disjointness, no independence needed.

By the hypothesis $\sum_{k \ge 1} m(E_k) < \infty$, the tail sums $\sum_{k \ge N} m(E_k)$ tend to 0 as $N \to \infty$ (tails of a convergent series).

Hence $0 \le m(\limsup E_n) \le \sum_{k \ge N} m(E_k) \to 0$, so $m(\limsup E_n) = 0$.

The proof is complete — this is the first Borel–Cantelli lemma, and problem 1's example is the special case $m(E_k) = 1/k^2$.

First, the product rigorously. By independence, $P\left(\bigcap_{k=N}^{M} E_k^c\right) = \prod_{k=N}^{M} P(E_k^c) = \prod_{k=N}^{M} (1 - \frac16) = (\frac56)^{M-N+1}$.

Taking logs, $\log\left((\frac56)^{M-N+1}\right) = (M-N+1)\,\log(\frac56)$, and since $\log(\frac56) < 0$ this tends to $-\infty$ as $M \to \infty$.

Hence $(\frac56)^{M-N+1} \to 0$ — the probability of "no 6 from toss $N$ through $M$" vanishes.

Now continuity of measure from above: $P$ is a probability measure, so $P \le 1 < \infty$ — the finite-measure hypothesis holds. The sets $\bigcap_{k=N}^{M} E_k^c$ decrease in $M$, so $P\left(\bigcap_{k \ge N} E_k^c\right) = \lim_{M \to \infty} P\left(\bigcap_{k=N}^{M} E_k^c\right) = \lim_{M \to \infty} (\frac56)^{M-N+1} = 0$ — for every $N$, "no 6 ever again from toss $N$ on" is a probability-0 event.

Since $(\limsup E_n)^c = \bigcup_{N \ge 1} \bigcap_{k \ge N} E_k^c$ (de Morgan), subadditivity gives $P((\limsup E_n)^c) \le \sum_{N \ge 1} P\left(\bigcap_{k \ge N} E_k^c\right) = \sum_{N \ge 1} 0 = 0$.

Hence $P(\limsup E_n) = 1$: a 6 occurs infinitely often, almost surely.

The same argument works in general — if the events are independent and $\sum p_n = \infty$, first separate the degenerate case: if $p_k = 1$ for some $k$, then $E_k$ occurs almost surely and the conclusion is trivial; otherwise assume $p_k < 1$ for all $k$, so $\log(1 - p_k)$ is defined, and $\log(1 - p_k) \le -p_k$ gives $\sum_{k=N}^{M} \log(1-p_k) \le -\sum_{k=N}^{M} p_k \to -\infty$, so the product tends to 0. That is the second Borel–Cantelli lemma.