5. Given that $f(x)$ is a function defined on $\mathbf{R}$, $f(1)=1$ and for any $x \in \mathbf{R}$, we have $f(x+5) \geqslant f(x)+5$, $f(x+1) \leqslant f(x)+1$. If $g(x)=f(x)+1-x$, then $g(2002)=$ $\qquad$ .
5. $g(2002)=1$ Hint: $f(x)=g(x)+x-1$, so $g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+$ 5, $g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1$. That is, $g(x+5) \geqslant g(x), g(x+1) \leqslant g(x)$. Therefore, $g(x) \leqslant$ $g(x+5) \leqslant g(x+4) \leqslant g(x+3) \leqslant g(x+2) \leqslant g(x+1) \leqslant g(x), g(x+1)=g(x), g(x)$ is a periodic function with a period of 1.