TITLE: Showing that asymptotic normality implies consistency.
QUESTION [6 upvotes]: In a statistics book i'm reading, it is postulated that asymptotic normality of an estimator implies consistency. That is
$$
\hat{\theta}_n \stackrel{as}{\sim} \mathcal{N}(\theta_0, \frac{1}{n}\sigma(\theta_0)) \Rightarrow P_{\theta_0}(|\hat{\theta}_n - \theta_0|>\epsilon ) \to 0
$$
when $n\to\infty$, for all $\theta_0 \in \Theta$ and $\epsilon>0$.
I am trying to prove this, but i can't seem to get a breakthrough. 
If anyone could shed some light on how this is proven, i would be very grateful.

REPLY [3 votes]: A more concise answer follows using the $O_p$ and $o_p$ notation for stochastic orders. Your premise is that 
$$\sqrt{n}(\theta_n-\theta_0)\overset{d}{\rightarrow}Z$$
where $Z$ is normally distributed with mean $0$ and variance $\sigma(\theta_0)$. Therefore
$$\theta_n-\theta_0=O_p\left(\frac{1}{\sqrt{n}}\right)\Rightarrow \theta_n-\theta_0= o_p(1)$$
hence $\theta_n-\theta_0\overset{p}{\rightarrow}0$, which finishes the proof.