Related with this question, we can see that $$f_{n}(x)=\sum_{k=0}^{2^{n}-1} \frac{k}{2^{n}}1_{I_{k}^{n}}(x)$$ We can see that this sequence converges uniformly to
$$ f(x) = x $$ in $[0, 1)$, my idea is prove by definition that
for every $\epsilon >0$, we can find $n_{0} \in \mathbb{N}$ such that for every $n \geq n_{0}$ and every $x_{0} \in [0, 1)$, we should have
$$|f_{n}(x)-f(x)|<\epsilon $$ Note that $$ |f_{n}(x)-f(x)|=\left |\left\lfloor 2^n x\right\rfloor -x \right| $$
But I don't understand the step to follow, in the post I am given an indication to continue with the proof and apparently it is easy but I don't understand what it means.
I ask again because I feel stuck. Any help would be appreciated!
From my comment, we have $$|f_{n}(x)-f(x)|= \frac{\left |\left\lfloor 2^n x\right\rfloor - 2^nx \right|}{2^n}.$$ Note that the fractional part of a real number $y$, i.e. $y - \lfloor y \rfloor$ lies in $[0, 1)$. Hence, $$|f_{n}(x)-f(x)| = \frac{\left |\left\lfloor 2^n x\right\rfloor - 2^nx \right|}{2^n} = \frac{2^n - \left\lfloor 2^n x\right\rfloor}{2^n} < \frac{1}{2^n}.$$ And, of course, as $n \to \infty$, $\frac{1}{2^n} \to 0$ (independently of $x$), hence $f_n \to f$ uniformly.