Originally Posted by

**moo** hello,

remember the definition (or property) of the continuity :

F is continuous $\displaystyle \leftrightarrow \forall \delta>0, \exists \epsilon>0,~ \forall x,~ |x-c|<\epsilon \rightarrow |f(x)-f(c)|<\delta$

(i inverted delta and epsilon from the usual definition so that it wouldn't confuse with the epsilon you're using in the text)

this means that the set : $\displaystyle \{\omega ~:~ |x_n(\omega)-a|< \epsilon \} \subseteq \{\omega ~:~ |g(x_n(\omega))-g(a)|< \delta \}$

one of the consequences of the kolmogorov axiomatic is that if $\displaystyle a \subseteq b$, then $\displaystyle \mathbb{p}(a) \leqslant \mathbb{p}(b)$

then use the second axiom of kolmogorov ($\displaystyle \mathbb{p}(a)\leqslant 1$) to prove that $\displaystyle \lim_{n \infty} \mathbb{p}(|g(x_n)-g(a)|<\delta)=1$

it should be easy from here (rofl)

yes.

I haven't been thinking very long on it, but my first idea would be, as above, to use the epsilon definition of $\displaystyle a_n \to a$ (nod)