Question on convergence of a random sequence after conditioning on a specific event

Consider a random process $\{X_n\}_{n=1}^\infty$ where each random variable is continuous. Assume that the sequence of random variables converges almost surely to $\alpha >0$ i.e., $$X_n \xrightarrow[n\rightarrow\infty]{\text{a.s.}} \alpha.$$ This implies that there exists $N$ such that $$\mathbb{P}(X_n >0, \text{ for all } n>N) =1.$$ I am trying to show that $$\mathbb{P}(X_n >0, \text{ for all } n>N | X_1>0 ,\dots,X_N>0) =1.$$ Here we assume that $\mathbb{P}(X_1>0,\dots,X_N >0 ) >0.$ My intuition is that this should hold because of the convergence result but I am not sure how to prove it formally. Any help would be really appreciated.

Note that $(A\cap B)\cup (A^c\cap B)=B$ where $A^c$ is the complementary event to $A$. Therefore $$\mathbb P(A\cap B)=\mathbb P(B)-\mathbb P(A^c\cap B).$$ Next, $\mathbb P(A^c\cap B)\leq \mathbb P(A^c)=0$ if $\mathbb P(A)=1$. Then $$\mathbb P(A\cap B)=\mathbb P(B)-\mathbb P(A^c\cap B)\geq \mathbb P(B).$$ Finally $$\mathbb P(A\mid B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} \geq \frac{\mathbb{P}(B)}{\mathbb{P}(B)} =1.$$
Small comment to the text of your question. Note that $N$ in your statement
This implies that there exists $N$ such that $$\mathbb{P}(X_n >0, \text{ for all } n>N) =1.$$
should depend on elementary event $\omega$: for almost every $\omega$ there exists $N=N(\omega)$ such that $$\mathbb{P}(X_n >0, \text{ for all } n>N(\omega)) =1.$$ Indeed, take $X_n=1+\frac{X}{n}$ where $X$ is standard normal. Then for each non-random $N$ $$\mathbb{P}(X_n >0, \text{ for all } n>N) = \mathbb P(X>-N)<1.$$
• Regarding the comment to the text of the question. Would your note be the same if $X_n \rightarrow \infty$ almost surely? – SpawnKilleR Sep 7 '18 at 20:23
• @SpawnKilleR Yes, it will be the same. Take $X_n=n+X$ and the last string remains unchanged. – NCh Sep 7 '18 at 23:43
Assuming that the sequence is i.i.d. and denoting $A = \{ X_n > 0: \forall n > N \}$ and $B = \{ X_1 > 0, \dots, X_N > 0 \}$ we have $$\mathbb{P}(A ~|~ B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = \frac{\mathbb{P}(A)\mathbb{P}(B)}{\mathbb{P}(B)} = \mathbb{P}(A) = 1,$$ as desired.