Example 8 A function $f(x)$ defined on $\mathbf{R}$ satisfies
$$
\begin{array}{l}
f(x)+f(1-x)=1, \\
f\left(\frac{x}{3}\right)=\frac{1}{2} f(x)(x \geqslant 0),
\end{array}
$$
$$
f\left(x_{1}\right) \leqslant f\left(x_{2}\right)\left(0 \leqslant x_{1}<x_{2} \leqslant 1\right) .
$$

Find the value of $f\left(\frac{1}{2011}\right)$.
【Analysis】This problem requires some observation and patient calculation.
Solution In equations (2) and (1), let $x=0$, we get
$$
f(0)=0, f(1)=1 \text {. }
$$

In equation (2), let $x=1$, we get
$$
f\left(\frac{1}{3}\right)=\frac{1}{2} f(1)=\frac{1}{2} \text {. }
$$

In equation (1), let $x=\frac{1}{2}$, we get $f\left(\frac{1}{2}\right)=\frac{1}{2}$.
Since $f(x)$ is a non-decreasing function, for any $x \in\left[\frac{1}{3}, \frac{1}{2}\right]$, we have $f(x)=\frac{1}{2}$.
$$
\begin{array}{l}
\text { Hence } f\left(\frac{1}{2011}\right)=f\left(\frac{1}{3} \times \frac{3}{2011}\right) \\
=\frac{1}{2} f\left(\frac{3}{2011}\right)=\frac{1}{2} f\left(\frac{1}{3} \times \frac{3^{2}}{2011}\right) \\
=\left(\frac{1}{2}\right)^{2} f\left(\frac{3^{2}}{2011}\right)=\cdots=\left(\frac{1}{2}\right)^{6} f\left(\frac{3^{6}}{2011}\right) \\
=\left(\frac{1}{2}\right)^{6} \times \frac{1}{2}=\frac{1}{128} .
\end{array}
$$