Example! Given that $f(x)$ is a function defined on $\mathbf{R}$, $f(1)=1$, and for any $x \in \mathbf{R}$, $f(x+5) \geqslant f(x)+5$, $f(x+1) \leqslant f(x)+1$. If $g(x)=f(x)+1-x$, find the value of $g(2002)$.
Analysis: The key to solving this problem is to determine the function using the inequality relationship.
Solution: From $g(x)=f(x)+1-x$ we get: $f(x)=g(x)+x-1$,
so $g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+5$,
$g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1$.
Therefore, $g(x+5) \geqslant g(x), g(x+1) \leqslant g(x), g(x) \leqslant g(x+5) \leqslant g(x+4) \leqslant g(x+3) \leqslant$ $g(x+2) \leqslant g(x+1) \leqslant g(x)$, so $g(x+1)=g(x)$, which means $g(x)$ is a periodic function with a period of 1. Given $g(1)=1$, thus $g(2002)=1$.