Consider the linear model
$y_i = x_i' \beta+u_i$ for $i=1,\ldots,n$
with $E(y_i \mid x_i)=x_i' \beta \iff E(u_i \mid x_i)=0$. Assume that the observations on $(y_i, x_i')$ are independent over $i=1,...,n$
The textbook claims that $E(u_i \mid x_i,\ldots,x_n)=E(u_i \mid x_i)$. Why is this? How does knowing that $(y_i, x_i')$ is independent from $(y_j, x_j')$ tell us that $E(u_i \mid x_j)=0$?
We're given that $(y_1,x_1'),\ldots,(y_n,x_n')$ are independent, and $u_i=y_i-x_i'\beta.$ Now let $X_{-i}$ be any collection of the $x_j$ that does not include $x_i,$ so $X_{-i}$ is independent of both $u_i$ and $x_i.$ Then we have the following probability density functions (assuming they all exist): $$\begin{align} f(u_i\mid x_i, X_{-i}) &={f(u_i,x_i,X_{-i})\over f(x_i, X_{-i})}\quad\text{by definition of the conditional p.d.f.}\\[2ex] &={f(u_i,x_i)f(X_{-i})\over f(x_i)f(X_{-i})}\quad\text{because $X_{-i}$ is independent of both $u_i$ and $x_i$}\\[2ex] &={f(u_i,x_i)\over f(x_i)}\\[2ex] &=f(u_i\mid x_i) \end{align}$$ Therefore, $\mathsf E(u_i\mid x_i,X_{-i})=\mathsf E(u_i\mid x_i).$
(The notation is sloppy, writing the same letter $f$ for all the different p.d.f.s, which are to be distinguished by the symbols used in their arguments; also, the same letters denote both random variables and their values.)