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.

  • 3
    $\begingroup$ When the underlying measure space is of finite measure, convergence a.e. implies convergence in measure. $\endgroup$
    – Ashok
    Mar 25, 2012 at 11:38
  • $\begingroup$ Too add a bit to what Ashok says, this fact can be derived from Egoroff's theorem. $\endgroup$
    – Alex R.
    Mar 26, 2012 at 0:39

2 Answers 2


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.


@Davide Giraudo's answer works quite well with the probability space, which has finite measure. And I am going to show a more general result, i.e.

$\textbf{Claim:}$ if the measurable space $(\Omega,\mathcal{F},\mu)$ has finite measure($\mu(\Omega)<\infty$), then the a.e. convergence implies convergence in measure.

which is suggested by @Ashok. Also, I will introduce another mode of convergence to help people understand the relationship between these modes of convergence(since I think it is hard to help the question owner after 8 years...).

Key observation is that

$X_n$ converges a.e. to $X$ is equivalent to(by definition) $$\mu(\mathop{\cap}\limits_{m=1}^{\infty}\mathop{\cup}\limits_{n=m}^{\infty}\{|X_n-X|\geq\epsilon\})=0,$$

while converges in measure is equivalent to(by definition as well) $$\lim\limits_{n\rightarrow\infty}\mu(\{|X_n-X|\geq \epsilon\})=0.$$

They are both true under the following statement: $$\lim\limits_{m\rightarrow\infty}\mu(\mathop{\cup}\limits_{n=m}^{\infty}\{|X_n-X|\geq \epsilon\})=0,$$

in all measurable space. And this kind of convergence is called $\textbf{almost uniform convergence}$, or a.u. convergence in short, since you can read from the above formula that for any $\epsilon>0$, there is some measurable set $A\in\mathcal{F}$ such that $\mu(A)<\epsilon$ and $X_n$ converges uniformly on $A$.

(Well, it may not be direct to "read" from it, but can be proved with moderate effort, and is a very good exercise for everyone :) Or you can go to Egorov's theorem for the proof--we only need a part of it to show the equivalence. However, I still recommend you to try on yourself.

Now, you can find: if the measure space is finite, then a.e. convergence is equivalent to a.u. convergence, since you can now pull the intersection out to be a limit, then implies convergence in measure. Q.E.D.

In general measure spaces, a.e. convergence and convergence in measure can be totally irrelevant:

  1. Let $X_n(x)=1(|x|>n)$ on $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$, then it converges a.e. but not in measure(Lebesgue measure);

  2. Let $X_n(x)=1(x\in[\frac{i}{2^k},\frac{i+1}{2^k}))$ be the famous "typewritter" function, where $n=2^k+i$, then it converges in measure but not a.e.


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