Haar functions are basis in $L^2[0,1]$ Define the Haar functions as $e_0^0=1$ and for $n\ge 1$, $k=1,\ldots,2^n$
$$
e_n^k(t)=\left\{ \begin{array}{ll}
         2^{\frac{n-1}{2}} & \mbox{if $x \in \big(\frac{K-1}{2^n},\frac{K}{2^n}\big)$};\\
        -2^{\frac{n-1}{2}} & \mbox{if $x \in \big(\frac{K}{2^n},\frac{K+1}{2^n}\big)$};\\
0 &\mbox{otherwise}
\end{array} \right.
$$
As part of showing that this is an orthonormal basis in $L^2[0,1]$, I need to show first that if $\langle f,e^k\rangle=0$ for each $k$ and $n$ then $\langle f,\chi_{[0,x]}\rangle = 0$ for every dyadic $x$. This would mean $f=0$. 
I was looking at this answer but I do not understand the construction he uses. Thanks in advance for your help.
 A: For some $n$ let $g \in L^2[0,1]$ be constant on each interval $[(k-1)/2^n,k/2^n)$ then $g = C +\sum_{k=1}^{2^n} b_k e_n^k$ where $b_k = 2^{-(n-1)/2} \sum_{m=0}^{k-1} g(m/2^n)$ and $C= \int_0^1 g(x)dx$
The Haar orthonormal basis is obtained by using this iteratively to show the $e_n^{2k}$ can be discarded
A: We can use $f_n^j := \frac1{2^{\frac{n-1}2}} e_n^j$ since they have the same linear span, denote it by $S$.
We shall prove by induction on $n$ that $$\chi_{\left(\frac{k-1}{2^n},\frac{k}{2^n}\right)} \in S, \quad \text{ for all } n \ge 0, 1 \le k \le 2^n.$$
Indeed, if $n=1$ then
$$\chi_{(0,1)} \stackrel{\text{a.e.}}{=} e_0^0$$
so $\chi_{(0,1)} \in S$.
Assume that for some $n \in \Bbb{N}$ we have
$$\chi_{\left(\frac{k-1}{2^n},\frac{k}{2^n}\right)} \in S, \quad \text{ for all } 1 \le k \le 2^n.$$
Pick an odd number $1 \le k \le 2^{n+1}-1$, or in other words $k=2r+1$ with $1 \le r \le 2^n-1$. We have
$$\chi_{\left(\frac{k-1}{2^{n+1}},\frac{k}{2^{n+1}}\right)} - \chi_{\left(\frac{k}{2^{n+1}},\frac{k+1}{2^{n+1}}\right)} = f_{n+1}^k\in S,$$
$$\chi_{\left(\frac{k-1}{2^{n+1}},\frac{k}{2^{n+1}}\right)} + \chi_{\left(\frac{k}{2^{n+1}},\frac{k+1}{2^{n+1}}\right)} \stackrel{\text{a.e.}}{=} \chi_{\left(\frac{k-1}{2^{n+1}},\frac{k+1}{2^{n+1}}\right)} = \chi_{\left(\frac{r}{2^n},\frac{r+1}{2^n}\right)} \in S$$
using the assumption. By taking $\frac12$ of the sum and difference, we get
$$\chi_{\left(\frac{k-1}{2^{n+1}},\frac{k}{2^{n+1}}\right)},\chi_{\left(\frac{k}{2^{n+1}},\frac{k+1}{2^{n+1}}\right)} \in S$$
which covers all of them so
$$\chi_{\left(\frac{k-1}{2^{n+1}},\frac{k}{2^{n+1}}\right)} \in S, \quad \text{ for all } 1 \le k \le 2^{n+1}$$
finishes the induction.
Now for any dyadic interval we have
$$\chi_{\left(0,\frac{k}{2^n}\right)} = \sum_{j=1}^{k-1} \chi_{\left(\frac{j-1}{2^n},\frac{j}{2^n}\right)} \in S$$
which proves your claim.
