A problem on a.s. convergence

I'm preparing for an exam and I came across this problem from old exams. I'm really clueless on how to solve it.

Consider a sequence of random variables $\{X_n\}_{n=1} ^\infty$ defined on the probability space $([0,1],B[0,1],\lambda)$ where $\lambda$ is Lebesgue measure. Define $X_n(\omega) = 1_{[1/2-(2n)^{-1}, 1/2+(2n)^{-1} ]}(\omega)$ for all $\omega \in [0,1]$. Prove that $X_n \xrightarrow{a.s.}0$ as $n \rightarrow \infty$.

• Borel-Cantelli lemma? – Cam.Davidson.Pilon Jan 8 '13 at 21:43
• @Cam: Just an aside: Note that Borel-Cantelli does not yield anything useful in this particular case, though a very minor modification to the function would. – cardinal Jan 8 '13 at 23:42

It is not hard to show that for every $\omega \neq \frac12$, $\exists N$ $\colon$ $X_n(\omega)=0$ for all $n \geq N$. Is it clear for you ? Then the conclusion is straightforward.
• Thanks! Can we say: $X_n(\omega) \rightarrow 0$ for all $\omega \in [0,1/2) \cup (1/2,1]$ and $\lambda( [0,1/2) \cup (1/2,1]) = 1$. Hence $X_n \xrightarrow{a.s.}0$ as $n \rightarrow \infty$ ? – user9292 Jan 8 '13 at 21:53
• @act00 Yes. Convergence for every $\omega \in [0,1/2) \cup (1/2,1]$ implies almost sure convergence because the Borelian set $[0,1/2) \cup (1/2,1]$ has full probability. Proving the convergence for $\omega \neq \frac12$ is not so straightforward but this is an elementary analysis exercise. – Stéphane Laurent Jan 8 '13 at 21:57
• (+1) The convergence proof is elementary: for $\omega\ne 1/2$, let $N$ be any integer greater than $1/|1-2\omega|$. The existence of such an $N$ is guaranteed by the Archimedean property of real numbers. An easy algebraic calculation establishes that $X_n(\omega)=0$ for $n\ge N$. – whuber Jan 8 '13 at 23:58