The Problem: Let $X$ be a random variable with cumulative distribution function $F_X$. Suppose that for all $t\in\mathbb R,F_X(t)=0$ or $F_X(t)=1$. Show that there exists a point $\varphi\in\mathbb R$ such that $P(X=\varphi)=1$. [Hint: Try the point $\varphi=\inf\{t\in\mathbb R:F_X(t)=1\}$.]
My Thoughts: Since $\lim\limits_{t\to-\infty}F_X(t)=0$, it follows that the set $\{t\in\mathbb R:F_X(t)=1\}$ is bounded below, so its infimum exists. Set $\varphi=\inf\{t\in\mathbb R:F_X(t)=1\}$.
For any $t_0\in\mathbb R$ with $\varphi<t_0$, we must have $F_X(t_0)=1$ since if $F_X(t_0)=0$, then $F_X(\varphi)=0$ by the monotonicity of $F_X$, which contradicts that $\varphi$ is the infimum of $\{t\in\mathbb R:F_X(t)=1\}$. Therefore, the right-continuity of the cumulative distribution function implies that $F_X(\varphi)=P(X\leq\varphi)=1.$
Next, for any $t_0<\varphi$, we have that $F_X(t_0)=0$, since otherwise $\varphi$ would not be the infimum of $\{t\in\mathbb R:F_X(t)=1\}.$ Therefore we have that
$$P(X<\varphi)=\lim\limits_{t\nearrow\varphi}F_X(t)=0.$$
Altogether we have
$$P(X=\varphi)=P(X\leq\varphi)-P(X<\varphi)=1.$$
What do you think about the proof above? Any feedback is much appreciated. Thank you for your time.