# Show that $P(B)>0$ where $B = \{\tau < T \text{ and } |X_t| \leq \theta -1 \text{ for all } 0 \leq t \leq \tau\}$ and $X$ is the sol of an Ito SDE

This question comes from a step in a proof in Mao's book on SDEs (page 120), where the author states that the below is true but he doesn't justify it.

Question: Consider the SDE on $$t \geq 0$$ $$dX_t = f(x_t,t)dt + g(X_t,t)dB_t, \quad X_0 = x_0 \in \mathbb R-{0}. \tag{1} \label{1}$$

Assume $$f$$ and $$g$$ satisfy the local Lipschitz and linear growth conditions so there exists a unique solution $$X$$ for SDE \eqref{1}. Define $$\tau := \inf \{t \geq 0 :X_t = 0\}$$ and assume $$P(\{\tau < \infty\})>0. \tag{2}$$

Show that we can find a pair of constants $$T>0$$ and $$\theta > 1$$ sufficiently large for $$P(B)>0,$$ where $$B = \{\tau < T \text{ and } |X_t| \leq \theta -1 \text{ for all } 0 \leq t \leq \tau\}.$$

My answer: I think that $$P(\tau < \infty) >0$$ implies that we can find a large enough $$T$$ such that $$P(\tau < T)>0.$$ So the original problem is equivalent to show that there exists a $$\theta > 1$$ such that $$P(\sup_{0 \leq t \leq \tau}|X_t|< \theta-1)>0. \tag{3} \label3$$ By contradiction, assume that \eqref{3} is false. Then, for all $$\theta >1,P(\sup_{0 \leq t \leq \tau}|X_t|\geq \theta -1)=1. \tag{4} \label4$$ But by Markov inequality $$P(\sup_{0 \leq t \leq \tau}|X_t|\geq \theta -1) \leq \frac{E(\sup_{0 \leq t \leq \tau}|X_t|)}{\theta -1} \leq \frac{E(\sup_{0 \leq t \leq T}|X_t|)}{\theta -1}.$$ Since $$f$$ and $$g$$ satisfied the linear growth condition, it can be shown (e.g. page 69 in the same book) that $$E(\sup_{0 \leq t \leq T}|X_t|) < C$$ for some constant $$C,$$ which doesn't depend on $$\theta.$$ Therefore, by taking $$\theta$$ such that $$\theta - 1 > C,$$ We have $$P(\sup_{0 \leq t \leq \tau}|X_t|\geq \theta -1)<1,$$ which contradicts \eqref{4}, so \eqref{3} is true.

Is my approach correct? Does $$P(\tau < \infty) >0$$ implies that we can find a large enough $$T$$ such that $$P(\tau < T)>0$$ (as I said at the beginning of my answer)?

$$P(\tau < T,\sup_{0 \leq t \leq \tau}|X_t|\leq \theta -1)=P(\tau < T)-P(\tau < T,\sup_{0 \leq t \leq \tau}|X_t|\geq \theta -1).$$
Let $$c_0:=P(\tau < T)$$, then we further take theta so large so that
$$P(\sup_{0 \leq t \leq \tau}|X_t|\geq \theta -1)\leq \frac{E(\sup_{0 \leq t \leq T}|X_t|)}{\theta -1}<\frac{c_{0}}{2},$$
$$P(\tau < T,\sup_{0 \leq t \leq \tau}|X_t|\leq \theta -1)\geq \frac{c_{0}}{2}.$$