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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2$\begingroup$ Possibly of interest: central-limit-theorem-versus-law-of-large-numbers. $\endgroup$– gung - Reinstate MonicaCommented Dec 6, 2013 at 23:59
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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.
