Example 1 (Adapted from the 2000 National High School Mathematics League, First Test) 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 $g(x)=f(x)+$ $1-x$, then $g(2009)=$ $\qquad$
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, $g(x) \leqslant g(x+5) \leqslant g(x+4) \leqslant g(x+2) \leqslant g(x+1) \leqslant g(x)$,
so $g(x+1)=g(x)$,
which means $g(x)$ is a periodic function with a period of 1. Given $g(1)=1$, hence $g(2009)=1$.