# Explanation of Durrett example 5.2.13

martingalesmeasure-theoryprobability theoryrandom walk

I am referring to Durrett's " Probability Theory and Examples". I am not including everything that's in the example as given in the book but only the relevant part for which I need explanation

Example 5.2.13:
Consider a simple symmetric random walk $$S_n = S_{n-1}+\zeta_n$$, where $$\zeta_n$$ takes values $$1$$ or $$-1$$ with probability $$\frac{1}{2}$$ each. Given $$S_0 = 1$$

Consider $$\tau = inf \{n: S_n = 0\}$$. Then, I know that $$X_n = S_{min(\tau,n)}$$ is a non-negative martingale, (so by martingale convergence theorem) $$X_n$$ converges to a random variable, call it $$X$$. Clearly, X is non-negative almost surely, and $$X$$ is finite almost surely.

The part I don't understand is how the author argues that $$X = 0$$ almost surely.

The author's argument as given in the book: "$$X_n$$ converges to a limit $$X$$, which is finite almost surely, that must be 0, since convergence to $$k > 0$$ is impossible (If $$X_n = k$$, then $$X_{n+1} = k+1$$ or $$k-1$$)"

Can anyone clarify how "(If $$X_n = k$$, then $$X_{n+1} = k+1$$ or $$k-1$$)" this implies $$X = 0$$ almost surely?

The only way a sequence of integers can converge is if that sequence is eventually constant. If $$\lim X_n = X$$ a.s. then for almost all $$\omega$$, we have $$X_n(\omega) = X_{n + 1}(\omega) = X_{n + 2}(\omega) = \dots$$ for large enough $$n$$ otherwise the limit would not exist.