I'm following Oksendal's book and during the construction of the Ito integral he uses the expectation of two different simple functions to demonstrate that where you choose the interior point of each interval of the partition is important for defining the correct integral. However, during that argument he chooses $\phi_2=\sum_{j=0}^{n-1} B_{t_{j+1}}\cdot\chi_{[t_j,t_{j+1})}$. Then he makes the following argument for some partition $\Pi_n=\{0=t_0<t_1<...<t_n=T\}$
$$\mathbb{E}(\int_0^T\phi_2\>dB_t)=\mathbb{E}[\sum_{j=0}^{n-1}B_{t_{j+1}}(B_{t_{j+1}}-B_{t_j})]=\sum_{j=0}^{n-1}\mathbb{E}[B_{t_{j+1}}(B_{t_{j+1}}-B_{t_j})].$$
This, I am fully ok with and understand, however it is this next line that I don't understand:
$$\sum_{j=0}^{n-1}\mathbb{E}[B_{t_{j+1}}(B_{t_{j+1}}-B_{t_j})]=\sum_{j=0}^{n-1}\mathbb{E}[(B_{t_{j+1}}-B_{t_j})^2].$$
Why is this the case?
Let $t>s$, then
$$E[B_t(B_t-B_s)]=E[(B_s + (B_t-B_s))(B_t - B_s)] \\ = E[B_s(B_t-B_s)] + E[(B_t-B_s)^2] \\ = E[B_s] E[B_t-B_s] + E[(B_t-B_s)^2] \\ = E[(B_t-B_s)^2].$$
The third equality basically follows by the Markov property. More precisely you can condition on $B_s$, use the Markov property for each fixed $B_s$, and then integrate over the possible choices of $B_s$; this procedure is called "applying the tower property of conditional expectation" or just "applying the tower property". The other equalities are straightforwardly justified.