Proof $\overline X_n \rightarrow \mu$ a.s. given $\sqrt{n}(\overline X_n - \mu) \rightarrow Z$ in distribution Given $X_1...X_n...$ i.i.d. random variables with $E|X_1|<\infty$ and $\sqrt{n}(\overline X_n - \mu) \rightarrow Z$ in law for a certain random variable Z and a certain real number $\mu$ (not necessarily equal to $EX_1$), I need to proof that $\overline X_n \rightarrow \mu$ a.s.
It is very similar to the central limit theorem but the problem is I don’t know if the variance is finite, so I cannot use it. Any hint on how can I approach this problem?
 A: As $\sqrt{n}(\overline X_n - \mu) \rightarrow Z$ in distribution and $\frac{1}{\sqrt{n}} \to 0$ hence by Slutsky's theorem we have $\frac{1}{\sqrt{n}}\sqrt{n}(\overline X_n - \mu) \rightarrow Z \cdot 0$ and $\overline X_n - \mu \to 0$ in distribution. As it is  convergence in distribution to $const$ then  $\overline X_n - \mu \to 0$ in probability. Hence $\overline X_n \to \mu  $ in probability.
From S.L.L.N. we have $\overline X_n \to EX_1$ a.s. Thus  $\overline X_n \to EX_1$ in probability.
As $\overline X_n \to \mu  $ in probability and $\overline X_n \to EX_1$ in probability we get that $\mu = EX_1$.
As $\overline X_n \to EX_1$ a.s. and $\mu = EX_1$. we get that $\overline X_n \to \mu$  a.s.
A: I'm working under the impression that $\overline{X}_{n} = n^{-1} \sum_{j = 1}^{n} X_{j}$.
Claim: Assume $\{X_{n}\}_{n \in \mathbb{N}}$ are i.i.d. random variables, $\mathbb{E}(|X_{1}|) < \infty$, and $Z$ is a random variable with $\mathbb{P}\{|Z| < \infty\} = 1$.  Let $\mu \in \mathbb{R}$.  If $\sqrt{n}(\overline{X}_{n} - \mu) \to Z$ in law, then $\mu = \mathbb{E}(X_{1})$.
As a consequence of the claim, the strong law of large numbers implies $\overline{X}_{n} \to \mu$.
Proof of the claim: By the strong law of large numbers and Egoroff's Theorem, there is an $\omega : [0,\infty) \to [0,\infty)$ and an event $\mathcal{C}$ such that $\mathbb{P}(\mathcal{C}) \geq 1/2$, $\lim_{\delta \to 0^{+}} \omega(\delta) = 0$, and
\begin{equation*}
|\overline{X}_{n} - \mathbb{E}(X_{1})| \leq \omega(n^{-1}) \quad \text{on} \, \, \mathcal{C}.
\end{equation*}
At the same time, since $\mathbb{P}\{Z < \infty\} = 1$, Prokhorov's Theorem implies that there is an $M > 0$ such that
\begin{equation*}
\mathbb{P}\{|\sqrt{n}(\overline{X}_{n} - \mu)| \geq M \} \leq 1/4.
\end{equation*}
It follows that, for each $n \in \mathbb{N}$, $\mathbb{P}(\{|\sqrt{n}(\overline{X}_{n} - \mu)| < M\} \cap \mathcal{C}) > 0$.  On that event, we have
\begin{equation*}
\sqrt{n}(\mathbb{E}(X_{1}) - \mu) \leq \sqrt{n} \omega(n^{-1}) + \sqrt{n}(\overline{X}_{n} - \mu) \leq \sqrt{n} \omega(n^{-1}) + M.
\end{equation*}
Since the extreme sides of the inequality are constants, we deduce that, for each $n \in \mathbb{N}$,
\begin{equation*}
\mathbb{E}(X_{1}) - \mu \leq \omega(n^{-1}) + M n^{-\frac{1}{2}}.
\end{equation*}
Sending $n \to \infty$, we find $\mathbb{E}(X_{1}) \leq \mu$.
A similar argument shows that $\mathbb{E}(X_{1}) \geq \mu$.
