I would like to prove that $\limsup_{t \to \infty} B_t = \infty$ a.s., where $B$ is a Brownian motion, by using the Fatou lemma.
My attempt
Fix $M>0$. Then $$P[\limsup_{t\to\infty} B_t>M]\geq\limsup_{t\to\infty}P[B_t>M]>0$$
I can conclude by noticing that $M$ is arbitrary, that the $\{\limsup_{t\to\infty} B_t>M\}$ is in the tail-$\sigma$ algebra and so it is $P$-trivial that
$$P[\limsup_{t\to\infty} B_t=\infty]=1$$
(Edit 1) To get the first inequality I proceed in this way
- Define $A_n:=\{\omega: B_n(\omega)>M\}$
- Define $A:=\limsup_{n\to\infty}A_n$
- Application of the Fatou's lemma for the '$\limsup$' to $$ \int\mathbb{1}_{A}dP=\int\mathbb{1}_{\limsup_{n\to\infty} A_n}dP{\color{red}=}\int\limsup_{n\to\infty}\mathbb{1}_{A_n}dP\geq\limsup_{n\to\infty}\int{\mathbb{1}_{A_n}}dP $$
Doubts:
- How can I pass from countable "n" to continuos "t"?
- Is the second (red) equality right? From this answer it seems so but the counter example given below seems to contradict it. (https://math.stackexchange.com/a/4728/1073326) Could someone explain it to me?
(Edit 2) As far as I understand from the answer gently given by Will the red equality is correct. My error is in the original set up of the sets $A_n$. So, can someone give me some insights on how to understand that these two sets are not the same: $\{\limsup_{n\to\infty} B_n>M\}$ and $\limsup_{n\to\infty}\{B_n>M\}$?
Thanks for the help.
How do you get $$P[\limsup_{t\to\infty} B_t>M]\geq\limsup_{t\to\infty}P[B_t>M]\quad?$$
Of course it is true for the brownian motion, because $\limsup_{t\to+\infty}B_t=+\infty$. But for any process $(B_t)_{t>0}$ that is not true. Take for instance $M=1$, $G$ a standard gaussian variable and $B_t=1+\frac Gt$. Then $P[\limsup_{t\to\infty} B_t>1]=0$ but $\limsup_{t\to\infty}P[B_t>1]=\frac12$.