Fatou lemma for limsup of a Brownian motion

Question

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:

1. How can I pass from countable "n" to continuos "t"?
2. 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.

Answer 1

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 $\limsup_{t\to\infty}P[B_t>1]=\frac12$.