If $E, F \in \mathcal{P}(X)$ such that $E = F$, then clearly $\mathcal{I}(E) = I_E = I_F = \mathcal{I}(F)$ which establishes injectivity of $\mathcal{I}$.

Next we must show surjectivity. Let $f : X \to \{ 0, 1 \}$ be a boolean function on $X$. Now $E : = f^{-1} \{ 1 \}$ is a set in $\mathcal{P}(X)$ such that $f$ attains value $1$ on every element in $E$ and value $0$ on every element in $X \setminus E$. Thus, $I_E = f$ by definition of an indicator function. Since we have established both injectivity and surjectivity, we are done. $\square$