consider a sequence $\{X_j\}$ of iid random variables where $X_j$ is in $L_1$ for each j Consider a sequence $\{X_j\}_{j\geq1}$ of iid random variables where $X_j$ is in $L_1$ for each $j$.
Suppose that the random variable $\frac{1}{\sqrt{n}} \sum_{j=1}^{n}(X_j-a)$
converges in law to the random variable $W$. 
Show that $\lim_{n\rightarrow \infty}\frac{1}{n} \sum_{j=1}^{n}X_j=a$
almost surely.
I know that, by the law of large number, $\frac{1}{n} \sum_{j=1}^{n}X_j$ converges almost surely to the expectation of $X_1$. But I don't know how to prove that $a$ is the expectation of $X_1$.
How can I solve it? Thank you for your help!
 A: Set
$$S_n := \sum_{j=1}^n (X_j-a).$$
By assumption, $S_n/\sqrt{n} \to W$ in distribution for some real-variabled random variable $W$. Fix $\delta, \epsilon>0$, and choose $R>0$ such that $\mathbb{P}(|W| \geq R) \leq \epsilon$. Since
$$\mathbb{P} \left( \left| \frac{S_n}{n} \right| > \delta \right) \leq \mathbb{P} \left( \left| \frac{S_n}{\sqrt{n}} \right| \geq R \right)$$
for $n \geq N=N(R)$ sufficently large, we find by the Portmanteau theorem
$$\limsup_{n \to \infty} \mathbb{P} \left( \left| \frac{S_n}{n} \right| > \delta \right) \leq \mathbb{P}(|W| \geq R) \leq \epsilon.$$
This shows that $S_n/n \to 0$ in probability. As
$$\mathbb{P} \left( \left| \frac{1}{n} \left( \sum_{j=1}^n X_j \right) - a \right|>\delta \right) = \mathbb{P} \left( \left| \frac{S_n}{n} \right|>\delta \right)$$
this means that $\frac{1}{n} \sum_{j=1}^n X_j \to a$ in probability. On the other hand, we have by the central limit theorem
$$\frac{1}{n} \sum_{j=1}^n X_j \to \mathbb{E}(X_1)$$
almost surely and hence in probability. Since limits in probability are unique (up to a null set), we get
$$a=\mathbb{E}(X_1).$$
Remark: The first part of this proof actually shows the following statement:

If $(Y_n)_{n \in \mathbb{N}}$ is a sequence which converges in distribution, then $\frac{Y_n}{\sqrt{n}} \to 0$ in probability.

