Given a function, $f$ be defined on $[0,1]$ by
$$f(x) = \begin{cases} 0 & x=0 \\ \frac1{2^{n-1}} & \frac1{2^n}\lt x\le\frac1{2^{n-1}}, n\in\Bbb N\end{cases}$$
If I take $n =1$, then $f(x) = 1$ and $\frac{1}{2} \lt x \le 1$
If $n =2$, $f(x) = \frac{1}{2}$ and $\frac{1}{2^{2}} \lt x \le \frac{1}{2}$
If $n=3$, $f(x) = \frac{1}{2^{2}}$ and $\frac{1}{2^{3}} \lt x \le \frac{1}{2^{2}}$
How should I find if the function is monotonically increasing or decreasing? Really don't know!
Take $x\in (2^{-m},2^{1-m}], y\in (2^{-n},2^{1-n}]$ for positive integers $m,n$ with $m\leq n$. Then $x\geq y$ and
$$f(x)=2^{1-m}=2^{1-n}2^{n-m}\geq 2^{1-n}=f(y),$$
so $f$ is monotonically increasing.