I'm looking at the constuction of the Brownian Motion given by Lévy-Ciesielki. We want to use Haar functions as basis of $L^2([0,1],\mathcal{B},\lambda)$. So on the n-th partition of $(0,1]$ $J_{k,n}:=2^{n/2}((k-1)2^{-n}]$ for $k=1,...,2^n$ (note that $J_{k,n}=J_{2k-1,n+1}\bigcup J_{2k,n+1}$ (disjoint union)we define the Haar function $f_{n,k}:=2^{n/2}(I_{2k-1,n+1}-I_{2k,n+1})$.
To check that the system $\{f_0,f_{n,k}\}_{n,k}$ is complete in order to prove that this is a basis of our function space,we have to check that $\forall f\in L^2:\langle f,f_{n,k}\rangle=0\Rightarrow f=0$.
Since $L^2$ is an Hilbert space, for any $f\in L^2$ we can express f through its Fourier series, $$\sum_k{\langle f,f_{n,k}\rangle f_{n,k}}$$. My problem is that in the lecture we tanke the partial sum of this Fourier series like follows. Set $\mathcal{B}_n=\sigma(J_{k,n+1};k=1,...,2^{n+1})$ as an increasing filtration to the Borel $\sigma$-algebra and so $$E_\lambda(f\mid\mathcal{B}_n)=\sum_{k=1}^{2^{n+1}}{2^{n+1}(\int f I_{J_{k,n+1}}d\lambda})I_{J_{k,n+1}}$$. I can't figure out how to get this partial sum. I tried the following:
$\langle f,f_{n,k}\rangle f_{n,k}(t)=\int_{[0,1]} f(s) 2^{(n+1/2)}(I_{2k-1,n+2}-I_{2k,n+2})(s)d\lambda(s)2^{(n+1)/2}(I_{2k-1,n+2}-I_{2k,n+2})(t)=[\int_{[0,1]} f(s) 2^{(n+1)/2}I_{2k-1,n+2}(s)d\lambda(s)-\int_{[0,1]} f(s) 2^{(n+1)/2}I_{2k,n+2}d\lambda(s)]2^{(n+1)/2}(I_{2k-1,n+2}-I_{2k,n+2})(t)=2^{n+1}[I_{2k-1,n+2}\int_{I_{2k-1,n+2}} f(s)d\lambda(s)-I_{2k-1,n+2}\int_{I_{2k,n+2}} f(s)d\lambda(s)-I_{2k,n+2}\int_{I_{2k-1,n+2}} f(s)d\lambda(s)+I_{2k,n+2}\int_{I_{2k,n+2}} f(s)d\lambda(s)]$
and with these signs I don't have chance to get what I wrote above for $E_\lambda(f\mid\mathcal{B}_n)$.