Talk:Famous Theorems of Mathematics/Law of large numbers

I don't follow the last step in this proof
The last step of this proof seem to suppose that if $$P(|\overline{X}_n-\mu|<\epsilon) \rightarrow 1$$ then $$\overline{X}_n \rightarrow \mu$$. Can someone explain this step for me? What about for example if it happens to be that $$X_i = \mu + \frac{1}{2}\ \!\epsilon, 1 \leq i \leq n,$$ then still $$P(|\overline{X}_n-\mu|<\epsilon) \rightarrow 1$$ but $$\overline{X}_n \rightarrow \mu + \frac{1}{2}\ \!\epsilon \neq \mu?$$


 * I think the confusion is coming about because in the proof we have that $$P(|\overline{X}_n-\mu|<\epsilon) \rightarrow 1$$ for every $$\epsilon$$, epsilon is fixed by the values of $$X_i$$. Thenub314 (talk) 10:00, 15 July 2009 (UTC)