10. Given that $f(x)$ is a function defined on $\mathbf{R}$. $f(1)=1$ and for any $x \in \mathbf{R}$, we have $f(x+5) \geqslant f(x)+5, \quad f(x+1) \leqslant f(x)+1$. If $g(x)=f(x)+1-x$, then $g(2002)=$ $\qquad$
10.1

From $g(x)=f(x)+1-x$ we get: $f(x)=g(x)+x-1$, so
$$
\begin{array}{l}
g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+5, \\
g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1 .
\end{array}
$$

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