Using result
R5362: Probability that a continuous random real number takes value in the rational numbers is zero and disjointness, we have
\begin{equation}
1
= \mathbb{P}(X \in \mathbb{R})
= \mathbb{P}(X \in \mathbb{Q} \text{ or } X \in \mathbb{I})
= \mathbb{P}(X \in \mathbb{Q}) + \mathbb{P}(X \in \mathbb{I})
= \mathbb{P}(X \in \mathbb{I})
\end{equation}
$\square$
