Try
Choose a point uniformly from [0,1]. The event that it lies in [0,1/2] has probability 1/2, while any one specified point has probability zero. These facts are compatible. Before calculating, distinguish an event, which is a set of outcomes, from an outcome itself. Infinite collections require more care than a finite list of equally likely possibilities.
Predict
Let A_n=[0,1−1/n] for positive integers n. Predict the union and its probability. The union is [0,1), which omits the endpoint 1 but still has probability one under the uniform law. Now predict why adding the probabilities of all singleton points is not a valid countable-additivity calculation.
Experiment
Record the event sequence.
| n | A_n endpoint | P(A_n) |
|---|
| 1 | 0 | 0 |
| 2 | 1/2 | 1/2 |
| 4 | 3/4 | 3/4 |
| 10 | 9/10 | 9/10 |
The events increase, and their probabilities approach one. For a decreasing example use B_n=[0,1/n]. Their intersection is the singleton zero, and their probabilities approach zero. Draw both sequences on the same unit interval. The limiting event is determined by membership across the sequence, not by simply substituting infinity into an endpoint.
Understand
A probability space consists of a sample space Ω, a collection of measurable events and a probability assignment P. The event collection is closed under complements and countable unions; this is called a sigma-algebra. Probability is nonnegative, assigns one to Ω, and is countably additive on disjoint events. This unit uses those rules without developing the full construction of measures.
Finite additivity gives complement and union rules. For overlapping A and B, P(A∪B)=P(A)+P(B)−P(A∩B). For a countable collection, subadditivity gives P(∪A_n)≤ΣP(A_n), even when the events overlap. The right side can exceed one; it is then merely a weak upper bound, not a probability value to report as exact.
If A_n increases, continuity from below gives P(∪A_n)=lim P(A_n). Decompose the union into the first event and the disjoint increments A_n minus A_(n−1); countable additivity turns the probability into the limit of partial sums. Decreasing-event continuity follows by applying that result to complements. Probability is finite, so the finiteness condition needed for general measures is automatically met.
Countability is essential. A countable union of zero-probability events still has probability zero, but an uncountable union can have probability one. The uniform interval consists of uncountably many singleton events, each with probability zero. There is no contradiction because countable additivity does not authorize summing over that uncountable family.
Probability zero need not mean an event is empty. Statements holding outside a zero-probability set hold almost surely. This distinction becomes important when conditional laws or random variables are defined only up to null sets. It does not mean that a model can ignore arbitrary inconvenient events; their probabilities must be justified within the model.
Countable Markov chains: returning is an infinite-horizon event
Let S be a finite or countable state set and P=(pᵢⱼ) a stochastic matrix. A discrete-time homogeneous Markov chain has conditional next-state law pᵢⱼ given its entire past ending in i. Starting at i, let Tᵢ⁺ be the first positive return time, fᵢ=Pᵢ(Tᵢ⁺<∞), and Nᵢ the total number of visits to i including time zero. A finite simulation that has not returned does not establish Tᵢ⁺=∞.
The Markov property also applies after an almost-surely finite stopping time: partition by the events T=n and the finite history through n, apply the ordinary Markov property there, and sum the resulting nonnegative probabilities over n and histories. The same partition proves restart on the event T<∞ even when T may be infinite. This proves the restart property used below; it does not assume future information is available at the stopping time.
Each completed return restarts the same return trial. Thus Pᵢ(Nᵢ≥k)=fᵢ^(k−1) for every positive integer k, by induction. The tail-sum identity and Tonelli give
EiNi=n=0∑∞pii(n)=k=1∑∞fik−1={(1−fi)−1,+∞,fi<1,fi=1. A state is recurrent when fᵢ=1 and transient otherwise. Recurrence means infinitely many visits almost surely, since Pᵢ(Nᵢ≥k)=1 for every k and the countable intersection still has probability one; transience gives finitely many visits almost surely because the displayed tail tends to zero. Expected visit count is infinite exactly in the recurrent case.
If i and j communicate, choose paths of lengths k,l with pᵢⱼ^(k)>0 and pⱼᵢ^(l)>0. The Chapman–Kolmogorov identity gives pᵢᵢ^(k+n+l)≥pᵢⱼ^(k)pⱼⱼ^(n)pⱼᵢ^(l). Summing over n shows that recurrence of j implies recurrence of i; reverse the roles for the converse. This proves recurrence/transience is a communicating-class property, even for an infinite state space. It does not claim all states in a reducible chain share a classification.
Positive and null recurrence are different from recurrence itself
For a recurrent state define mᵢ=EᵢTᵢ⁺. It is positive recurrent if mᵢ<∞ and null recurrent if mᵢ=∞. A finite irreducible chain is positive recurrent: from every state choose a path to i of length at most L and positive probability; finiteness gives a uniform lower bound δ>0 on reaching i within L steps. Repeated blocks bound the hitting-time tail by (1−δ)^k, a summable geometric bound. One additional step handles the first positive return.
The finite-state conclusion does not extend automatically to a countable chain. For a simple symmetric walk on the integers, starting at 1, the probability of hitting m before 0 is 1/m. To verify this, the stopped walk in the integer interval from 0 to m is bounded and exits almost surely (each block of m steps has uniformly positive exit probability), so the elementary bounded stopping identity gives E exit=1. The two possible exit values are 0,m. Let m tend to infinity: avoiding 0 forever requires reaching every m, apart from an event of probability zero corresponding to remaining in a finite interval forever. Therefore the walk hits 0 almost surely. Starting at−1 is symmetric, so 0 is recurrent.
Its expected return time is infinite. If the hitting time of 0 from 1 had finite expectation, bounded increments would allow stopping the martingale Sₙ and passing to the limit in expectation, giving E S_T=1. But S_T=0, a contradiction. The optional-stopping extension proves exactly that finite-mean-time justification. Translation and irreducibility give null recurrence throughout this walk. In contrast, a deterministic drift Xₙ=X₀+n never returns and is transient; the two-state alternating chain returns in exactly two steps and is positive recurrent, despite being periodic.
For a positive-recurrent state i in an irreducible chain, define expected occupation counts during one return cycle by νⱼ=EᵢΣₙ₌₀^(Tᵢ⁺−1)1(Xₙ=j). Then Σνⱼ=mᵢ and νP=ν: conditioning one step ahead replaces visits at times 0,…,T−1 by times 1,…,T, whose endpoint discrepancy is zero because both endpoints are i. Tonelli justifies the nonnegative sums. Thus πⱼ=νⱼ/mᵢ is a stationary probability distribution, and πᵢ=1/mᵢ. This proves construction from positive recurrence. We can also prove the converse and uniqueness without treating a guessed invariant vector as sufficient evidence.
Prove the stationary converse by reversing paths
Suppose an irreducible countable chain has a stationary probability π. Every πⱼ is positive: take a state i with positive mass and a positive-probability path from i to j, and use π=πPᵏ. Under the stationary initial law, let Bₙ be the event that time n is a visit to j with no later visit. The Bₙ are disjoint and each has probability πⱼ(1−fⱼ). A countable sum cannot exceed 1, so fⱼ=1. Thus every state is recurrent.
Moreover, from any recurrent state j, any reachable i is hit almost surely. Choose a simple path from j to i with probability δ>0 and no return to j before i. Each excursion from j has probability at least δ of reaching i; restarting at its successive returns bounds the chance of failing in k excursions by(1−δ)ᵏ. This tends to zero.
Define the reversed matrix pⱼᵢ=πᵢpᵢⱼ/πⱼ. Stationarity makes its row sums 1 and π is also stationary for P. Reversed positive paths show it is irreducible, so the same recurrence/hitting argument applies to P. For fixed n, decompose stationary paths ending at j according to their last visit to i among times 0,…,n−1. If that visit occurred k steps earlier, its contribution is πᵢPᵢ(Xₖ=j,Tᵢ⁺≥k). The remainder Rₙⱼ consists of paths with no such visit. Reversing each finite path gives Rₙⱼ=πⱼ times the P-probability from j of avoiding i for the next n steps. This tends to zero by the preceding hitting property (using positive return when j=i).
Consequently πⱼ=πᵢΣₖ₌₁^∞Pᵢ(Xₖ=j,Tᵢ⁺≥k). For j≠i this sum is exactly νⱼ, the occupation before the first return; for j=i it is the total return probability 1, also νᵢ. Hence πⱼ=πᵢνⱼ. Summing nonnegative terms over j gives 1=πᵢmᵢ, proving finite mean return and the formula πᵢ=1/mᵢ. The occupation counts depend only on P and i, so normalization fixes the unique stationary probability. Applying this argument to the stationary law constructed from any positive-recurrent state also proves that positive recurrence is a class property in an irreducible chain. Periodicity is irrelevant to this existence theorem; convergence of Pⁿ to π requires further hypotheses.
Classify a changed chain before inspecting the answer
On the nonnegative integers, let 0 be absorbing, and from n≥1 move to n−1 with probability 1/2 and n+1 with probability 1/2. Is state 1 recurrent because the unrestricted symmetric walk is recurrent? No. The chain hits 0 almost surely by the preceding finite-interval argument. Thereafter it cannot return to 1. More directly, from 1 the first step goes to 0 with probability 1/2 and never returns, so f₁≤1/2<1. State 0 is positive recurrent with return time 1. All positive states communicate with each other but not with 0, so their transience does not contradict the class theorem. Review criteria: state the altered transition at 0, distinguish communication from reachability, and use an actual positive nonreturn probability rather than a finite simulated path.
Primary comparator: MIT18.445 Spring 2015 lectures 12–13, checked 30 October 2026. These original arguments prove return classification, the positive-recurrence/stationary-probability equivalence and uniqueness for irreducible countable chains. Mixing rates and continuous-time explosion are not included.
Apply
Prove continuity from below with the increment construction, naming where disjointness is used. Then explain the uniform-singleton example without claiming that probabilities fail to add for finite disjoint events. A correct argument keeps the size of the event collection visible.
Continue
Next restrict the sample space using observed information. Conditional probability changes the reference event, and independence is a specific relationship between those restrictions.
