Show that if $X:[0,T]\to\mathbb R$ is continuous, then $\inf\left\{t\in\overline I:|X(t)|\geε\right\}\le t$ iff $|X(s)|\geε$ for some $s\in[0,t]$ Let


*

*$T>0$

*$I:=(0,T]$

*$X:\overline I\to\mathbb R$ be continuous with $X(0)=0$

*$\varepsilon>0$

*$\tau:=\inf\left\{t\in\overline I:|X(t)|\ge\varepsilon\right\}$

*$t\in\overline I$



How can we show that $\tau\le t$ if and only if $|X(s)|\ge\varepsilon$ for some $s\in[0,t]$?

Clearly, the "if" part is trivial. However, I don't know how we need to argue in the "only if" part. For example, why isn't it possible that $\tau=t$, $|X(s)|<\varepsilon$ for all $s\in[0,t]$ and $|X(s)|\ge\varepsilon$ for all $s\in(t,T]$?
 A: It suffices to show that $|X(\tau)| = \epsilon$.
Observe that $\tau \in I$.  Suppose $|X(\tau)|<\epsilon$.  Take $\eta = \frac12(\epsilon - X(\tau))$.  The continuity of $X$ gives a $\delta$-neighbourhood of $\tau$ so that $X(t') \in (X(\tau)-\eta, X(\tau)+\eta) \subseteq (X(\tau)-\eta, \epsilon)$ for all $t' \in (\tau-\delta, \tau+\delta)$.  This contradicts the definition of $\tau$ as a greatest lower bound of $\{|X| \ge \epsilon\}$ (because any $t' \in (\tau, \tau + \delta)$ is a greater lower bound for $\{|X| \ge \epsilon\}$).

For this question, we're done with $|X(\tau)| \ge \epsilon$.  (It's given that $\tau \le t$, so $s = \tau$ satisfies the question.)  The rest is left for exercise.
Remarks: In probability-theory, $\tau$ is called the hitting time of the event $\{|X| \ge \epsilon\}$, where $\Omega = [0,T]$ equipped with the usual Borel $\sigma$-algebra.  Then $X$ is a random variable starting from $0$.  It's quite intuitive that $|X(\tau)| = \epsilon$ due to the continuity of $X$.
A: Let $A=\{t\in [0,T]: |X(t)|\geq  \varepsilon\}.$ 
If $t\in [0,T]$ and $|X(s)|< \varepsilon$ for all $s\in [0,t]$ then  $A\subset [t,T].$ So if $A$ is not empty then $\tau =\inf A\geq \inf [t,T]=t .$
If $t\in [0,T]$ and $|X(s)| \geq \varepsilon$ for some $s\in [0,t]$ then $s\in A$ so  $\tau =\inf A\leq s\leq t.$ 
