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As in subject, is there a sequence of random variables $\left \{ X_{n} \right \}_{n\in\mathbb{N}}$ $\mathbb{P} (\lim_{n \rightarrow \infty} X_{n} = 0) = 0.5$?

I say yes, because take $A=\left \{-1, 0 \right \}$ and $X_{n}(\omega)=\omega^{n}$.

Then $\lim_{n \rightarrow \infty} X_{n}(0) = 0$ and the limit of $X_{n}(-1)$ not exists.

So $\mathbb{P}(\left \{ \omega\in A: \lim_{n\rightarrow \infty} X_n =0\right \}) = \mathbb{P}(\left \{ 0 \right \}) = 0.5 $

But I have a feeling that I am wrong. What about Kolmogorov's 0-1 theorem? What about independence of $X_n$?

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    Are you requiring the sequence $(X_n)$ to be independent or are you not? If you are, the answer is "no", for the reason you say. If you are not, the answer is "yes", as tons of examples show. – 2017-01-30
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    I was interested if there exists such a sequence if we do not know anything about dependence. – 2017-01-30
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    Then see third sentence of my comment. – 2017-01-30
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    To see that without the independence assumption the answer is trivial, consider the elementary sequence: $P(X_1 = 0) = 0.5, P(X_1 = 1) = 0.5$, then $X_n = X_1\forall n>1$ (a.s.). Then clearly $P(\lim_{n \to \infty}X_n = 0) = P(X_1 = 0) = 0.5$ as desired. – 2017-01-30

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Your example is correct. But, as you correctly state, due to Kolmogoroff, your variables cannot be independent. And, of course, they aren't, as can for example be seen via $$ P (X_{n+1}=0 \mid X_n = 0) = 1 \ne \frac 12 = P (X_{n+1}=0)$$

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    You must've refer $1\neq 1/2 = P(X_{n+1}=0)$ – 2017-01-30