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Let $X_n$ be an independent sequence of centered, integrable real random variables on $(\Omega, \mathcal{A}, P).$ Show that if $X_n$ obeys the weak law of large numbers, then the sequence $n^{-1}X_n$ converges in probability to $0$.

I need to show that $$P\left(\left|\frac{1}{n}X_n\right|\ge \epsilon\right)\to 0$$ for all $\epsilon>0$. From the hypothesis, we have $$P\left(\left|\frac{1}{n}\sum_{i=1}^nX_i\right|\ge \epsilon\right)\to 0$$ for all $\epsilon>0$. How can I get the above from using this? I would greatly appreciate any help.

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    $\begingroup$ Hint. You have $\frac{1}{n}X_n = \bar{X}_n - \frac{n-1}{n}\bar{X}_{n-1}$, where $\bar{X}_n = \frac{1}{n}(X_1 + \cdots + X_n)$. $\endgroup$ – Sangchul Lee Jan 25 '17 at 1:59
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Let $\displaystyle S_n = \sum_{k=1}^n X_k$. Notice that $\displaystyle \frac{S_n}{n} = \displaystyle \frac{S_{n-1}}{n-1}\frac{n-1}{n} + \frac{X_n}{n}$, hence, $$ \left\{\omega\in\Omega: \left|\frac{X_n}{n}\right|>\epsilon\right\}\subset \left\{\omega\in\Omega:\left|\frac{S_n}{n}\right|>\frac{\epsilon}{2}\right\}\cup\left\{\omega\in\Omega:\left|\frac{S_{n-1}}{n-1}\right|>\frac{n}{n-1}\frac{\epsilon}{2}\right\} $$, therefore, $$ \left\{\omega\in\Omega: \left|\frac{X_n}{n}\right|>\epsilon\right\}\subset \left\{\omega\in\Omega:\left|\frac{S_n}{n}\right|>\frac{\epsilon}{2}\right\}\cup\left\{\omega\in\Omega:\left|\frac{S_{n-1}}{n-1}\right|>\frac{\epsilon}{2}\right\}. $$ Using union bound and lettting $n\to\infty$, we get $$ \limsup_{n\to\infty}\mathbb{P}\left(\left\{\omega\in\Omega: \left|\frac{X_n}{n}\right|>\epsilon\right\}\right)\leq \lim_{n\to\infty}\mathbb{P}\left(\left\{\omega\in\Omega:\left|\frac{S_{n}}{n}\right|>\frac{\epsilon}{2}\right\}\right) +\lim_{n\to\infty}\mathbb{P}\left(\left\{\omega\in\Omega:\left|\frac{S_{n-1}}{n-1}\right|>\frac{\epsilon}{2}\right\}\right)=0 $$ hence, $\displaystyle\frac{S_n}{n}\to 0$ in probability.

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