Proof of Weak Law of Large Numbers

For a sequence of i.i.d. random variable $$(X_n)$$ with $$\Bbb{E}X=m$$, $$\overline{X}_n=(X_1+...+X_n)/n$$, then $$\overline{X}_n\stackrel{P}{\rightarrow}m$$.

Since $$(X_n)$$ are i.i.d., $$(X_n)$$ induce the same probability measure $$\mu$$ on $$\Bbb{R}$$. So $$\Bbb{P}\{|\overline{X}_n-m|>\epsilon\}=\int_{\{|\frac{x_1+...+x_n}{n}-m|>\epsilon\}}\Bbb{1}(\mathrm{d}\mu)^n$$ where $$(\mathrm{d}\mu)^n$$ is the product measure of $$\mu$$. I want to know if we can prove weak law of large numbers by this representation. Any help will be appreciated.

• You should use the Chebyshev inequality. First, you might want to compute $Var\left(\overline{X}_n\right)$. This would of course assume $X_1$ has finite variance. – Michael Apr 15 at 3:27
• @Michael Yes, sir! I know the standard proof, first consider the case of finite variation and use Chebyshev inequality, then using truncation to approximate the general case. But I want to know if my attempt is useful. – Xin Fu Apr 15 at 3:38
• I don't see how your integral helps to identify concentration properties. The key idea in all proofs of the law of large numbers is to compute the variance of either $\overline{X}_n$ or its truncated version. Somewhere you will want to use the fact that $E[(X_i-m)(X_j-m)]=0$ (or its truncated version $E[(\tilde{X}_i-m_i)(\tilde{X}_j-m_j)]=0$). – Michael Apr 15 at 3:41