Example 3 Given that $f(x)$ is a function defined on $\mathbf{R}$, $f(1)=1$. And for any $x \in \mathbf{R}$, there are $f(x+5) \geqslant f(x)+5, f(x+1) \leqslant f(x)+1$. If $g(x)=f(x)+1-x$, find $g(2002)$.
(2002 National High School League Question)
Given $g(x)=f(x)+1-x$, we get $f(x)=g(x)+x-1$, so
$$
\begin{array}{l}
g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+5 \\
g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1
\end{array}
$$

which means
$$
g(x+5) \geqslant g(x), g(x+1) \leqslant g(x)
$$

Therefore, $g(x) \leqslant g(x+5) \leqslant g(x+4) \leqslant g(x+3) \leqslant g(x+2) \leqslant g(x+1) \leqslant g(x)$
$$
g(x+1)=g(x)
$$

Thus, $g(x)$ is a periodic function with a period of 1. Given $g(1)=1$, hence $g(2002)=1$.