Root-$n$ consistency of an estimator implies its variance is $O(n^{-1})$

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(This is from a claim on page 20 of Hall's 1992 The Bootstrap and the Edgeworth Expansion.)

Suppose we observe i.i.d. data $X_1,\ldots,X_n$ from some unknown CDF $F$. Let $\theta_0=\theta(F)$ be a parameter that can be computed from the CDF and $\hat\theta$ an estimator.

Hall then mentions: "should $\hat\theta$ be $\sqrt{n}$-consistent for $\theta_0$, so that $E(\hat\theta-\theta_0)^2=O(n^{-1})$ ..."

Why is this true?

Could you please in particular provide a rigorous argument (you can make further assumptions if necessary)? My interest in this study is to see how such an argument can be constructed.

My progress so far is that I understand the claim intuitively, as follows: the $\sqrt{n}$-consistency of $\hat\theta$ means there is some fixed positive number $V$ such that $$ \sqrt{n}(\hat\theta-\theta_0)\approx N(0,V)\implies E[(\hat\theta-\theta_0)^2]\approx\frac{1}{n}V=O(n^{-1}). $$

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By the continuous mapping theorem, $n(\hat \theta-\theta_0)^2 \xrightarrow{\mathcal L} \mathcal N(0,1)^2 $.

If $n(\hat \theta-\theta_0)^2$ is uniformly integrable then $E(n(\hat \theta-\theta_0)^2) \to 1$ and the bound easily follows.

I'm not sure there's much you can say without the uniform integrability assumption. The weak Fatou's lemma yields $1\leq \liminf_n E(n(\hat \theta-\theta_0)^2)$.