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Let $(X_n : n ≥ 1)$ be the sequence of random variables on the standard unit-interval probability space, as shown below.

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I am trying to determine whether this sequence converges almost surely. My reasoning is as follows: for a fixed $\omega$, the limit of $X(\omega)$ (as $n \to \infty$) is equal to $0$ (this is because as $n \to \infty$, the values $X(\omega)$ are $0$ more and more frequently and $1$ less and less frequently -- it is as though we have the following sequence of numbers: $0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, ...$, which converges to $0$). And for this reason, the sequence converges almost surely.

However, the book argues that the limit of $X(\omega)$ (for a fixed $\omega$, as $n \to \infty$) does not exist, and therefore, the sequence does not converge almost surely. What is the flaw in my reasoning?

StopReadingThisUsername
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1 Answers1

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For any fixed $\omega$ the sequence alternates between zero and one infinitely often.

It is true that there are "more" zeros than ones in the sequence. But both occur infinitely often, thus the sequence does not converge.

Simon
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