Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $\mathbb F=(\mathcal F)_{t\ge 0}$ be a filtration on $(\Omega,\mathcal A)$
- $B=(B_t)_{t\ge 0}$ be an $\mathbb F$-adapted Brownian motion with respect to $\mathbb F$
Let $H=(H_t)_{t\ge 0}$ be $\mathbb F$-adapted, locally bounded and of the form $$H_t(\omega)=\sum_{i=1}^nH_{t_{i-1}}(\omega)1_{(t_{i-1},t_i]}(t)\;\;\;\text{for all }\Omega\times[0,\infty)\;$$ for some $0=t_0<\ldots<t_n$. The Itō integral of $H$ with respect to $B$ is defined as $$I_\infty^B(H):=\sum_{i=1}^nH_{t_{i-1}}\left(B_{t_i}-B_{t_{i-1}}\right)\;.\tag{1}$$
Let $\mathcal E$ be the set of all such $H$. Many people state, that $$I_\infty^B: \mathcal E\to\mathcal L^2(\operatorname P)$$ is a linear mapping. But, begin a mapping would mean, that if we choose another partition $(t_0',\ldots,t_{n'}')$ such that $$H_t(\omega)=\sum_{i=1}^{n'}H_{t'_{i-1}}(\omega)1_{(t'_{i-1},t'_i]}(t)\;\;\;\text{for all }\Omega\times[0,\infty)\;,$$ we would need to have $$\sum_{i=1}^nH_{t_{i-1}}\left(B_{t_i}-B_{t_{i-1}}\right)=\sum_{i=1}^{n'}H_{t'_{i-1}}\left(B_{t'_i}-B_{t'_{i-1}}\right)\;,$$ but I absolutely don't understand why this should be the case.