I'm reading the book Hilbert Space Methods in Probability and Statistical Inference by Small and McLeish.
There (Def. 3.2.4) independence is defines as follows
Let $\mathbf{B}$ and $\mathbf{C}$ be two sets of random variables. The sets $\mathbf{B}$ and $\mathbf{C}$ are said to be independent if $$\langle \mathbf{x} - \mathbb{E}(\mathbf{x} )\mathbf{1}, \mathbf{y} - \mathbb{E}(\mathbf{y} )\mathbf{1}\rangle =0$$ for all $\mathbf{x} \in ps(B)$ and all $\mathbf{y} \in ps(C)$. In particular, if $\mathbf{B}=\{\mathbf{x}\}$ and $\mathbf{C}=\{\mathbf{y}\}$, then we say that $\mathbf{x}$ and $\mathbf{y}$ are independent. Random variables $\{\mathbf{x}_{\alpha}\}$ are said to be mutually independent if any two disjoint subcollections of the random variables are independent. Similarly events $\{\mathbf{A}_{\alpha}\}$ are said to be mutually independent if their corresponding indicators are mutually independent.
The authors claim, that this definition immediately implies $$P\left( \bigwedge_{i=1}^{n}A_{i} \right)=\prod\limits_{i=1}^{n}P(A_{i}) $$ for mutually independent events $A_{1},A_{2},\ldots, A_{n}$.
While I think the proof for arbitrary $n$ follows easily by induction, I'm stuck in proving the result for $n=2$.
From the definition, two events $A$ and $B$ are mutually independent if: $$\begin{align} \langle \mathbf{1}_{A} - \mathbb{E}(\mathbf{1}_{A} )\mathbf{1}, \mathbf{1}_{B} \mathbb{E}(\mathbf{1}_{B} )\mathbf{1}\rangle &= \langle \mathbf{1}_{A} , \mathbf{1}_{B} \rangle - \mathbb{E}(\mathbf{1}_{B})\langle\mathbf{1}_{A} ,\mathbf{1} \rangle - \mathbb{E}(\mathbf{1}_{A})\langle\mathbf{1}_{B} ,\mathbf{1} \rangle + \mathbb{E}(\mathbf{1}_{A})\mathbb{E}(\mathbf{1}_{B})\langle \mathbf{1} ,\mathbf{1}\rangle\\ &= \langle \mathbf{1}_{A} , \mathbf{1}_{B} \rangle - 2\mathbb{E}(\mathbf{1}_{A})\mathbb{E}(\mathbf{1}_{B}) + \mathbb{E}(\mathbf{1}_{A})\mathbb{E}(\mathbf{1}_{B})\\ &= \langle \mathbf{1}_{A} , \mathbf{1}_{B} \rangle - \mathbb{E}(\mathbf{1}_{A})\mathbb{E}(\mathbf{1}_{B})\\ &= 0 \end{align}$$ Therefore $$\langle \mathbf{1}_{A} , \mathbf{1}_{B} \rangle =\mathbb{E}(\mathbf{1}_{A})\mathbb{E}(\mathbf{1}_{B})$$ implies independence.
Furthermore, also be definition, $P(A):=\mathbb{E}(\mathbf{1}_{A})$. The last equality is therefore equivalent to $$\langle \mathbf{1}_{A} , \mathbf{1}_{B} \rangle = P(\mathbf{A})P(\mathbf{B})$$
However to complete the proof $P(A \bigwedge B)=\langle \mathbf{1}_{A} , \mathbf{1}_{B} \rangle $ must hold. But by definition $$P\left(A \bigwedge B\right)=\mathbb{E}(\mathbf{1}_{A\bigwedge B})=\langle \mathbf{1}_{A}\mathbf{1}_{B},\mathbf{1} \rangle$$.
If we consider the Hilbert Space $L^{2}([0,1])$ this follows from the definition of $\langle f,g\rangle :=\int_{[0,1]}f(x)g(x)dx$ since $$ \langle \mathbf{1}_{A}\mathbf{1}_{B},\mathbf{1}\rangle = \int_{[0,1]}\mathbf{1}_{A}(x)\mathbf{1}_{B}(x)\mathbf{1}_{[0,1]}(x)dx = \int_{[0,1]}\mathbf{1}_{A}(x)\mathbf{1}_{B}(x)(x)dx = \langle \mathbf{1}_{A},\mathbf{1}_{B}\rangle $$
But I do not see why $$\langle\mathbf{1}_{A}\mathbf{1}_{B},\mathbf{1} \rangle=\langle \mathbf{1}_{A} , \mathbf{1}_{B}\rangle$$ should hold in general? I assume I'm missing some property of inner products.