3. For any subset $U$ of the set of real numbers $\mathbf{R}$, we define the function $f_{U}(x)=\left\{\begin{array}{ll}1, & x \in U \\ 0, & x \notin U\end{array}\right.$. If $A, B$ are two subsets of the set of real numbers $\mathbf{R}$, then $f_{A}(x)+f_{B}(x) \equiv 1$ if and only if
For $\forall x \in \mathbf{R}, f_{A}(x)+f_{B}(x) \equiv 1 \Rightarrow x \in A \bigcup B$, and $x \notin A \cap B$; conversely, for $\forall x \in \mathbf{R}, x \in A \cup B$, and $x \notin A \cap B \Rightarrow f_{A}(x)+f_{B}(x) \equiv 1$. Therefore, the necessary and sufficient condition for $f_{A}(x)+f_{B}(x) \equiv 1$ is $A \cup B=\mathbf{R}$, and $A \cap B=\varnothing$.