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Assuming that the distribution has finite variance (a condition not required for the LLN), then doesn't the LLN follow from the CLT?

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WLLN, yes.

Here is a general claim: Suppose $\{ f_n \}$, $f$, and $g$ are random variables, and

$$ \sqrt{n} (f_n - f) \stackrel{d}{\mapsto} g. $$

Let's say the CDF of $g$ is continuous everywhere. Then $f_n \rightarrow f$ in probability. This is because $\sqrt{n} (f_n - f)$ is bounded in probability/uniformly tight.

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