# Thread: lim inf contained in lim sup

1. ## lim inf contained in lim sup

Hey,

$(A_{n})_{n \geq 1}$ a sequence of subsets of $\mathcal{A}$, a $\sigma$-field on $\Omega$.

With
$\lim_{n \to \infty} \inf A_{n} = \bigcup_{n=1}^{\infty} \bigcap_{m\geq n} A_{m}$
and
$\lim_{n \to \infty} \sup A_{n} = \bigcap_{n=1}^{\infty}\bigcup_{m \geq n} A_{m}$

show that $\lim_{n \to \infty} \inf A_{n} \subset \lim_{n \to \infty} \sup A_{n}$

Hints on how to start?

2. ## Re: lim inf contained in lim sup

Let $x\in\liminf_{n\to +\infty} A_n$. We can find $n_0$ such that for $n\geq n_0$, $x\in A_n$. Now, you have to show that for all $k\in\mathbb{N}$, $x\in\bigcup_{j\geq k}A_j$.
"To be in $\liminf_{n\to +\infty} A_n$" means "to be in all $A_n$ for $n$ large enough" whereas "To be in $\limsup_{n\to +\infty} A_n$" means "to be in infinitely many $A_n$".

3. ## Re: lim inf contained in lim sup

$x \in \lim_{n \to \infty} \inf A_{n} \Rightarrow \exists n_0 : x \in A_{n} \forall n \geq n_{0}$

If $x \in \bigcup_{j \geq k} A_{j}$ but $x \notin \bigcup_{j \geq k+1} A_{j}$ this means $x \notin \lim_{n \to \infty} \inf A_{n}$ which is a contradiction.

And clearly $x \in \bigcup_{j \geq 1} A_{j}$, whence

$x \in \lim_{n \to \infty} \sup A_{n}$

$\lim_{n \to \infty} \inf A_{n} \subset \lim_{n \to \infty} \sup A_n{n}$

Is this ok?

4. ## Re: lim inf contained in lim sup

Yes, it works.