If $\sqrt{n}(T_n - \theta) \stackrel{d}{\to} T$, show that $T_n \stackrel{p}{\to} \theta$ I think we're supposed to use the slutsky's theorem for this but I'm not exactly sure how to proceed. The other thing I thought of was whether or not the delta method would be of use here anywhere but I'm stuck.
I appreciate any form of help. Thank you. 
 A: This can be proven by following the definitions. We need that for any positive $\epsilon$ and $\delta$ there exists an $n_0 \in \mathbb{N}$ such that 
$$ \mathbb{P}(|T_n - \theta| \ge \epsilon) \le \delta, \ \forall \ n \ge n_0$$
We can use the continuity of the probability measure to find an $m \ge 0$ such that $$\mathbb{P}( \ |T| \ge \epsilon \cdot \sqrt{m} \ ) \le \delta/2.$$
Then we can compute for $n \ge m$
$$ \mathbb{P}(|T_n - \theta| \ge \epsilon) = \mathbb{P}( \ \sqrt{n} \cdot|T_n - \theta| \ge \epsilon \cdot \sqrt{n} \ ) \le \mathbb{P}( \ \sqrt{n} \cdot|T_n - \theta| \ge \epsilon \cdot \sqrt{m} \ )$$
Now by the convergence in distribution there exists an $n_0$ greater or equal to $m$ such that 
$$
 \mathbb{P}( \ \sqrt{n} \cdot|T_n - \theta| \ge \epsilon \cdot \sqrt{m} \ ) \le \mathbb{P}(\ |T| \ge \epsilon \cdot \sqrt{m} \ ) + \delta/2, \ \forall \ n \ge n_0.
$$
So that together with the first inequality we have that:
$$ \mathbb{P}(|T_n - \theta| \ge \epsilon) \le \delta/2 + \delta/2 = \delta.$$
