# Strong law of large numbers implies weak law

I'm reading Achim Klenke's textbook on probability theory, and I've hit a road block in his proof that the strong law of large numbers (SLLN) implies the weak law of large numbers (WLLN).

Let $(X_n)_{n \in \mathbb N}$ be a sequence of real integrable random variables, and let $\widetilde S_n = \sum_{i=1}^n (X_i - \mathbb E[X_i])$, where $\mathbb E[X]$ is the expectation of the random variable $X$. The weak and strong laws of large numbers are:

• WLLN: $\displaystyle \lim_{n \to \infty} \mathbb P\left[ \left| \frac 1 n \widetilde S_n\right| > \epsilon\right]=0$.
• SLLN: $\displaystyle \mathbb P\left[ \limsup_{n \to \infty} \left| \frac 1 n \widetilde S_n\right|=0\right] = 1$.

For $\epsilon > 0$, Klenke defines the sets $A_n^\epsilon = \left\{ \left| \frac 1 n \widetilde S_n \right| > \epsilon\right\}$ and $A = \left\{ \limsup_{n \to \infty} \left| \frac 1 n \widetilde S_n \right| > 0 \right\}$, and states that $$A = \mathop\bigcup_{m \in \mathbb N} \limsup_{n \to \infty} A_n^{1/m},$$ and thus $\mathbb P\left[ \limsup_{n \to \infty} A_n^\epsilon\right] = 0$ for $\epsilon>0$. Then, by Fatou's lemma, \begin{align} \limsup_{n \to \infty} \mathbb P\left[A_n^\epsilon\right] &= 1-\liminf_{n \to \infty} \mathbb E\left[ \mathbb 1_{\left(A_n^\epsilon\right)^c}\right] \leq 1 - \mathbb E\left[\liminf_{n \to \infty} \mathbb 1_{\left(A_n^\epsilon\right)^c}\right] = \mathbb E\left[ \limsup_{n \to \infty} \mathbb 1_{A_n^\epsilon}\right] = 0. \end{align} My question: I understand this entire argument, except for the first equality in the above equation. Equivalently, what I'm trying to figure out is, why is $\limsup \mathbb P\left[A_n^\epsilon\right] = 1-\liminf \mathbb P\left[ \left(A_n^\epsilon\right)^c\right]$? Is this some general measure theory result that I'm not seeing?

## 1 Answer

I think this may just be a real-analysis thing. In particular, I think that the author is using the fact that $$\lim\sup x_n = - (\lim\inf - x_n).$$ This follows from the definition of $\sup$ and $\inf$.

In particular, put $x_n = \mathbb{P}[A^\epsilon_n]$, and note that $\mathbb{P}[A^\epsilon_n] = 1 - \mathbb{P}[(A^\epsilon_n)^c]$.

Then we can rewrite the right-hand side as $$1 - \lim\inf (1 - x_n) = 1 - 1 - \lim\inf - x_n = -\lim\inf - x_n.$$ We can then apply the result above.

• Yes, that's exactly what's being used. Took me a second but that's the gist. Thanks! – D Ford Nov 26 '17 at 17:55