# Show that $P(X_t=0) = 1$ and $P(sup_{0 \leq t \leq 1}X_t = 1) = 1$

Let $([0,1],B[0,1],\lambda)$ and let $\left\{X_t, 0 \leq t \leq 1\right\}$ be a family of random variables defined as: $X_t(\omega)=\begin{cases} 1 \text{ for } \mathbb\omega=t & \\ 0 \text{ otherwise} & \end{cases}$

Show that $P(X_t=0) = 1$ and $P(sup_{0 \leq t \leq 1}X_t = 1) = 1$.

I don't understand how to solve it. Someone can give me a detailed answer ?

Some hints: For $P(X_t = 0)$, let's recall that this is just $P(\{\omega : X_t(\omega) = 0\})$. Is there an easier way of writing the set $\{\omega : X_t(\omega) = 0\}$? Since your measure is just the Lebesgue measure, the probability of this event is just the Lebesgue measure of that set.
Try to think of what the function $\sup_{0 \leq t \leq 1} X_t$ is equal to. Fix a value of $\omega \in [0,1]$; then what is $\sup_{0 \leq t \leq 1} X_t(\omega)$?
• Thanks for the hint. I've understood why $P(X_t = 0) = 1$. I've still trouble with the sup: I think the $sup_{0 \leq\ t \leq 1}X_t(\omega) = 1$, but isn't the Lebesgue measure of 1 = zero? – jo_jo Sep 24 '17 at 17:33
• Well, what about $X_\omega(\omega)$? – Marcus M Sep 24 '17 at 19:22
• @jo_no, Don't think about interpreting it just yet. Think about how each $X_t(\omega)$ is a function of $\omega$; what value does $X_\omega(\omega)$ take? – Marcus M Sep 25 '17 at 17:31