Let $(X_j)_j\geq 1$ be i.i.d. with $X_j$ in $L_1$. Suppose $\frac{1}{\sqrt{n}}\sum_{j=1}^n(X_j-\nu)\rightarrow Z$ in distribution where $Z$ is a random variable. What we need to show is
$\lim\limits_{n\rightarrow \infty}\frac{1}{n} \sum\limits_{j=1}^n X_j = \nu$ almost surely.
The solution I found is to show this by using characteristic function. https://www.springer.com/cda/content/document/cda_downloaddocument/ProbEssSolutions.pdf?SGWID=0-0-45-1313337-p2054396.
Is there another way to show the convergence?
Lemma If $\xi_n \to \xi$ and $\xi_n+c_n \to \xi '$ in distribution then $\{c_n\}$ is bounded. Proof of the lemma: there exists $M$ such that $P\{|\xi_n| >M\} <1/2$ and $P\{|\xi_n+c_n| >M\} <1/2$ for $n$ sufficiently large. [ This is because these probabilities converge to $P\{|\xi|>M\}$ and $P\{|\xi '|>M\}$ respectively and these probabilities can be made small by choosing $M$ to be large number at which the limiting distributions are continuous. If you are familiar with tightness the you would already know this]. Now suppose $|c_n|>2M$ for infinitely many $n$. Then either $|\xi_n| >M$ or $|\xi_n+c_n| >M$ with probability 1. But the probability of this event is $<1/2+1/2$. This contradiction proves the Lemma. Now let $\xi_n=\frac {X_1+...+X_n-n\nu} {\sqrt n}$ Let $c_n=n\nu-nm$ where $m=EX_1$. $\xi_n \to Z$ in distribution by hypothesis and $\xi_n+c_n \to \xi '$ in distribution for some normal random variable $\xi '$ by CLT. It follows that $\{n\nu -nm\}$ is bounded. But this means $\nu =m$. Now the conclusion follows by SLLN.