# prove $P\{\int_{0}^{\infty}f(W_s)ds=\infty\}=1$

where $$W_s$$ is standard Brownian motion and $$f:\mathbb{R}\to [0,\infty)$$ is Borel-measurable function satisfies that $$\{y\in \mathbb{R}:f(y)>0\}$$ has positive Lebsgue measure.

Is $$\int_{0}^{\infty}f(W_s)ds$$ a Brownian motion? can I use the recurrent property of the Brownian motion to prove this?

• when you say that f has positive measure, do you mean that $\int_0^\infty f(s)\text{d}s > 0$? Jun 19 at 13:20
• @Davius I think that means the set that satisfies $f(y)>0$ is not a zero measurement set. Jun 19 at 13:25

We need to show that $$\int_0^T f(B_t) dt\to\infty, T\to\infty.$$ I will use the occupation density formula (UPD see a simpler argument below): $$\int_0^T f(B_t) dt = \int_{\mathbb{R}} f(x) L^x_T(B) dx,$$ where $$L_T^x(B) = \lim_{\varepsilon \to 0} \frac{1}{2\varepsilon}\int_0^T \mathbf{1}_{[x-\varepsilon,x+\varepsilon]}(B_t) dt$$ is the local time of $$B$$ at $$x$$ on $$[0,T]$$. Since $$B$$ is recurrent, for each $$x\in\mathbb{R}$$, $$L_T^x(B)\to +\infty$$, $$T\to +\infty$$. Therefore, by the Fatou lemma, $$\liminf_{T\to\infty} \int_0^T f(B_t) dt \ge \int_{\mathbb {R}} f(x)\liminf_{T\to\infty} L_T^x(B_t) dx = +\infty.$$
In fact, one can show (e.g. using the self-similarity of Brownian motion) the following convergence in distribution (if $$f$$ is not integrable, then the convergence is to infinity): $$\frac{1}{\sqrt{T}} \int_0^T f(B_t)dt \overset{d}{\rightarrow} \int_{\mathbb R} f(x) dx \cdot L_1^0(B)\overset{d}{=} \int_{\mathbb R} f(x) dx \cdot |B_1| , T\to \infty.$$
Here is a simpler argument. From the assumption it follows that there are some $$a such that $$\int_a^b f(x) dx >0.\tag{1}$$
Define $$\tau_0= 0$$, $$\sigma_n = \min\{t\ge \tau_{n-1}: B_t = a\},\quad \tau_n = \min\{t\ge \sigma_n: B_t = b\}, \quad n\ge 1.$$ From the recurrence it follows that, almost surely, $$\sigma_n$$ and $$\tau_n$$ are well defined for all $$n\ge 1$$.
Now $$\int_0^\infty f(B_t) dt\ge \sum_{n=1}^\infty \int_{\sigma_n}^{\tau_n} f(B_t) dt=: \sum_{n=1}^\infty I_n$$. From the strong Markov property of Brownian motion it follows that the random variables $$I_n$$ are iid, and from (1), that they are positive. Therefore, $$\sum_{n=1}^\infty I_n=\infty$$ a.s. (e.g. by SLLN).