Brownian motion with a stopping time

Let $$x \geq 0,c<0,$$ and a Brownian motion $$(W_u)_{u}.$$ Let $$T:=\inf\{u \geq 0, B_u +cu\geq x\}.$$

It follows that $$Y:=\sup_{u \geq 0}(B_u+cu) \in ]0,\infty[.$$

We want to verify that $$\{Y \geq x\} \subset (T<\infty).$$

Supposing that $$Y(w) > x$$ then the result follows.

But what if $$Y(w)=x,$$ I can't see how to deduce it ? Do we need to use the continuity of the BM?

Suppose $$T=\infty$$. Then $$B_u+cu for all $$u$$ and $$B_{u_n}+cu_n \to x$$ for some sequenece $$u_n \to \infty$$ (because $$Y=x$$). But the $$\frac {B_{u_n}} {u_n}+c \to 0$$. This is a contradiction becasue $$\frac {B_{u_n}} {u_n} \to 0$$ (a.s.)and $$c <0$$.
• If $Y(w)=x,$ can we claim that in this case $Y(w) \in \{B_u+cu,u \geq 0\}$ ?
• No, you cannot. But there is a sequence in the set converging to $Y(\omega)$. Commented May 14, 2022 at 0:18
• If there is a sequence $u_n$ such that $Y(w)$ is the limit of $B_{u_n}+cu_n,$ it possible to prove that $u_n$ is convergent (or has subsequence) ? (We are not supposing that $T(w)=\infty$)
• If $T(\omega) <\infty$ then $(u_n)$ can converge to a finite limit. But if $T(\omega)=\infty$ the $u_n$ must tend to $\infty$. @Riro Commented May 14, 2022 at 4:56