10. 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 $g(x)=f(x)+1-x$, then $g(2002)$ $=$ . $\qquad$
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10.【Analysis and Solution】From $g(x)=f(x)+1-x$ we get: $f(x)=g(x)+x-1$, so
$$
g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+5, g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1 \text {. }
$$

That is, $g(x+5) \geqslant g(x), g(x+1) \leqslant g(x) . \therefore g(x) \leqslant g(x+5) \leqslant g(x+4) \leqslant g(x+3) \leqslant g(x+2)$ $\leqslant g(x+1) \leqslant g(x) \ldots g(x+1)=g(x)$.
That is, $g(x)$ is a periodic function with a period of 1, and since $g(1)=1$, we have $g(2002)=1$.