Let $(h_n)_{n\in\mathbb{N}}$ be the Haar orthonormal basis. In our script we have the following equation as part of a proof for the existence of a Brownian Motion:
$\sum_{n=0}^{\infty} \langle h_n , \mathbb{1}_{[0,t]} \rangle \langle h_n, \mathbb{1}_{[0,s]} \rangle = \langle \mathbb{1}_{[0,t]}, \mathbb{1}_{[0,s]} \rangle $
Where the brackets represent the standard $L_2$ scalar product (i.e. integral over $\Omega$) and $s < t$ are in an intervall $[0, T]\subset \mathbb{R}$. I know of this equation for general orthonormal basis:
$x = \sum_{n=0}^{\infty} \langle x, h_n \rangle h_n$
But I still can't figure out what happened here. This might be an error in the script but I can't tell. Can someone clarify please?
Choose $x = 1_{[0, t]}$ and scalarly mutiply your "known" equation by $x$: $$ \langle x, x \rangle = \left \langle \sum_{n = 1}^\infty \langle x_n, h_n \rangle h_n, x\right \rangle $$ We use continuity of the scalar product: $$ \left \langle \sum_{n = 1}^\infty \langle x_n, h_n \rangle h_n, x\right \rangle = \left \langle \lim_{m \rightarrow \infty}\sum_{n = 1}^m \langle x_n, h_n \rangle h_n, x\right \rangle = \lim_{m \rightarrow \infty} \left \langle \sum_{n = 1}^m \langle x_n, h_n \rangle h_n, x \right \rangle = \lim_{m \rightarrow \infty} \sum_{n = 1}^m \langle \langle x_n, h_n \rangle h_n, x \rangle = \lim_{m \rightarrow \infty}\sum_{n = 1}^m \langle x_n, h_n \rangle \langle x_n, h_n \rangle = \sum_{n = 1}^\infty \langle x_n, h_n \rangle \langle x_n, h_n \rangle $$ (The last sum exists in $\mathbb{R}$ according to Bessel's inequality.)