Let
For $n\in \mathbb{N}$ and $k\in \{0, 1, 2, ..., 2^{n}-1 \}$ is defined
$$I_{k}^{n}=\left[\frac{k}{2^{n}}, \frac{k+1}{2^{n}}\right)$$
and $f_{n}:[0, 1) \rightarrow \mathbb{R}$ is defined by
$$f_{n}(x)=\sum_{k=0}^{2^{n}-1} \frac{k}{2^{n}}1_{I_{k}^{n}}(x)$$
Where for $A\subseteq \mathbb{R}$, the function $1_{A}$ is the characteristic of $A$ such that $1_{A}(x)=1$ if $x\in A$ and $1 _{A}(x)=0$ if $x\notin A$.
The idea is to prove that $f_n \rightarrow f$ in $[0, 1)$. So, let's fix $x_0 \in [0, 1)$. We need to show by definition that
$$ \lim_{{n \to \infty}} f_n(x_0) = f(x_0) = x_0. $$
Given $\epsilon > 0$, we must find $n_0 \in \mathbb{N}$ such that if $n \geq n_0$, then
$$ \left| \sum_{{k=0}}^{{2^n-1}} \frac{k}{2^n}1_{I_k^n}(x_0) - x_0 \right| < \epsilon. $$
Now, let's consider this by cases:
If $x_0 = 0$, then
$$ \lim_{{n \to \infty}} f_n(x_0) = \lim_{{n \to \infty}} \sum_{{k=0}}^{{2^n-1}} \frac{k}{2^n}1_{I_k^n}(x_0) = 0 = x_0. $$
So,
$$\left| \sum_{{k=0}}^{{2^n-1}} \frac{k}{2^n}1_{I_k^n}(x_0) - x_0 \right| = 0 < \epsilon. $$ And for any choice of $n_{0}$ it would be enough.
Now, I am unable to determine this limit for $0 < x < 1$. The behavior of the sequence of functions in that interval is somewhat unusual. I attempted to find some values in that interval, as described in the following image:
I don't see how to determine the limit of $f_{n}$ at those points, any suggestions? I appreciate it!
Let $x_0 \in [0,1)$, for all $n \in \mathbb{N}$, $(I_k^n)_{k \in [0..2^{n-1}]}$ are disjoint and form a partition of $[0,1)$.
So for all $n$, there exist a unique $k$ such that $f_n(x_0)=k/2^n$. And for all $n$ and all $k$ the length of $I_k^n$ is $\frac{k+1}{2^n}- \frac{k}{2^n} = \frac{1}{2^n}$.
Let $\epsilon >0$,
As $lim_{n \to \infty} \frac{1}{2^n}= 0$, there exist $n_0 \in \mathbb{N}$ such that for all $n \geq n_0$, $\frac{1}{2^{n}}< \epsilon$.
Let $n \geq n_0$, there exist a unique $k$ such that $f_{n}(x_0)=k/2^{n}$, $x_0 \in [\frac{k}{2^n},\frac{k+1}{2^n})$, and $|f_n(x_0)-x_0|= |\frac{k}{2^n} - x_0 | \leq |\frac{k+1}{2^n}- \frac{k}{2^n}| = \frac{1}{2^n}$, and we chose $n$ such that $\frac{1}{2^n} < \epsilon$, so $|f_n(x_0)-x_0|< \epsilon$
Edit: one can remark that $n_0$ does not depend on $x_0$, thus you have also uniform convergence.