Compare convergence modes using event bounds

Try

At stage n an approximation may fail on an event A_n. If the failure probabilities sum to a finite number, the approximation fails only finitely often with probability one. This is stronger than merely saying each probability tends to zero. Summability controls the entire infinite tail, even when the events are dependent.

Predict

Predict whether P(A_n)→0 alone implies that only finitely many failures occur. The typewriter sequence of dyadic intervals disproves that claim. Predict whether a divergent probability sum forces infinitely many events. It does under independence, but not for arbitrary dependence: repeatedly using the same event gives a divergent sum without probability-one occurrence.

Experiment

List the first two full levels of dyadic intervals on [0,1).

Level	Intervals visited	Probability of each indicator being one
1	[0,1/2), [1/2,1)	1/2
2	Four consecutive intervals of length 1/4	1/4
3	Eight consecutive intervals of length 1/8	1/8

Pick x=1/3 and mark its one interval at every level. Its indicators contain infinitely many ones and zeros, even though each late indicator has small probability of being one. The half-open convention handles endpoints consistently.

Understand

Convergence almost surely means convergence outside a null set of outcomes. Convergence in probability means P(abs (X_n−X)>ε)→0 for every ε>0. Convergence in Lp means E abs (X_n−X)^p→0. All random variables and the chosen limit are assumed measurable.

Almost-sure convergence implies convergence in probability by dominated convergence applied to error indicators bounded by one. Lp convergence implies convergence in probability by Markov applied to abs (X_n−X)^p. The typewriter example shows that convergence in probability need not imply almost-sure convergence of the whole sequence. Concentrating spikes show that almost-sure convergence need not imply L1 convergence without additional control.

For the first Borel–Cantelli lemma, write the infinitely-often event as the intersection of the tail unions. The probability of each tail union is at most the tail sum of P(A_n). If the series is finite, these tails tend to zero, proving that the infinitely-often event is null. No independence was used.

For the second lemma assume independence and divergence of the probability sum. For fixed N, the probability of avoiding every event from N through m is the product of 1−P(A_n), at most exp (−ΣP(A_n)). This tends to zero. Continuity above implies that avoiding all events after N has probability zero. Take the countable union over N to conclude that only finitely many occurrences has probability zero.

Finally, convergence in probability always has an almost-surely convergent subsequence. Choose increasing n_k with P(abs (X_(n_k)−X)>2^-k)<2^-k. Borel–Cantelli shows that almost every outcome eventually satisfies every selected tolerance, proving the subsequence claim. This extraction is a key ingredient in Vitali’s theorem and many compactness arguments.

Apply

Construct a dependency counterexample for the second lemma by setting every A_n equal to one event of probability 1/2. The sum diverges, but the infinitely-often event still has probability 1/2. Then explain why this counterexample does not challenge the first lemma: its summability hypothesis is absent.

Continue

Always identify both the convergence mode and the quantifier order. A subsequence certificate is not a whole-sequence certificate. In the laws-of-large-numbers lesson, summable maximal error estimates will bridge that gap and prove an almost-sure result for every sufficiently late sample size.

Continue to Define conditional expectation by integral tests.