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The strong law implies the weak law (although there seem to be variations on exactly what hypothesis are assumed for the strong law vs the weak law which probably make this not true depending on your definition of the strong law and weak law). The weak law states convergence in probability whereas the strong law states convergence almost surely.

Why does the strong law imply the weak law? In real analysis convergence in measure and convergence almost everywhere don't imply each other. Can someone explain what are the subtle meanings behind the difference convergences and what they mean for the law of large numbers.

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When the underlying measure space is of finite measure, convergence a.e. implies convergence in measure. –  Ashok Mar 25 '12 at 11:38
    
Too add a bit to what Ashok says, this fact can be derived from Egoroff's theorem. –  Alex R. Mar 26 '12 at 0:39
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Th fact that the strong law of large numbers implies the weak law of large numbers in contains in the following property:

If $(\Omega,\mathcal F,\Bbb P)$ is a probability space and $\{X_n\}$ a sequence of real valued random variables with converges almost everywhere to $X$, it converges in probability to $X$.

To see that, we can consider $X_n\to 0$ and $X_n\geq 0$, replacing $X_n$ by $|X_n-X|$ if necessary. The set $$C:=\bigcap_{p\geq 1}\bigcup_{n\geq 1}\bigcap_{k\geq n}\{\omega\in\Omega,X_k\leq \frac 1p\},$$ has by hypothesis, measure $1$. Fix $\varepsilon>0$ and $p_0$ with $p_0^{-1}\leq \varepsilon$. We have \begin{align} \limsup_k\Bbb P\{X_k\geq \varepsilon\}&\leq \limsup_k\Bbb P\{X_k\geq p_0^{-1}\}\\ &\leq\inf_{n\in\Bbb N}\Bbb P\left(\bigcup_{k\geq n}\{X_k\geq p_0^{—1}\}\right)\\ &\leq \Bbb P\left(\bigcap_{n\in\Bbb N}\bigcup_{k\geq n}\{X_k\geq p_0^{—1}\}\right)\\ &\leq \Bbb P(\Omega\setminus C)=0, \end{align}
which gives the wanted result.

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