Given that $h_i(p,u) = \frac{de(p,u)}{p_i}$, why does this equality hold? $p^0$ is a vector $(p^0_1, p^0_2, p^0_3....)$ and $p^1$ is a vector $(p^1_1, p^1_2,....)$ . Shouldn't the upper bound and the lower bound switch in the first term? And why is the second term true?
Source: MGW Microeconomic Theory
You are correct, and the solution that you attached have some minor errors.
Since it is a line integral, the following should be true.
$$ e(p^0, u^1) - e(p_1^1, p_2^0, p_3^0, \cdots, p_L^0, u^1) = \int_{p_1^1}^{p_1^0}h(p_1, p_2^0, p_3^0, \cdots, p_L^0, u^1)dp_1 $$ , or $$ = -\int_{p_1^0}^{p_1^1}h(p_1, p_2^0, p_3^0, \cdots, p_L^0, u^1)dp_1 $$
However, note that the solution reaches the goal anyway because it makes the same mistake for the CV.
Instead, I will give you my corrected solution to the problem.
Note that $$ \begin{align} EV(p^0, p^1, w) & = e(p^0, u^1) - e(p^1, u^1) \\ & = e(p^0, u^1) - e(p_1^1, p_2^0, p_3^0, \cdots, p_L^0, u^1) + e(p_1^1, p_2^0, p_3^0, \cdots, p_L^0, u^1) - e(p^1, u^1) \\ & = \int_{p_1^1}^{p_1^0}h(p_1, p_2^0, p_3^0, \cdots, p_L^0, u^1)dp_1 + \int_{p_2^1}^{p_2^0}h(p_1^1, p_2, p_3^0, \cdots, p_L^0, u^1)dp_2 \end{align} $$
Moreover, $$ \begin{align} CV(p^0, p^1, w) & = e(p^0, u^0) - e(p^1, u^0) \\ & = e(p^0, u^0) - e(p_1^1, p_2^0, p_3^0, \cdots, p_L^0, u^0) + e(p_1^1, p_2^0, p_3^0, \cdots, p_L^0, u^0) - e(p^1, u^0) \\ & = \int_{p_1^1}^{p_1^0}h(p_1, p_2^0, p_3^0, \cdots, p_L^0, u^0)dp_1 + \int_{p_2^1}^{p_2^0}h(p_1^1, p_2, p_3^0, \cdots, p_L^0, u^0)dp_2 \end{align} $$
Now, under the no wealth effect, note that $$ h(p_1, p_2^0, p_3^0, \cdots, p_L^0, u^1) = h(p_1, p_2^0, p_3^0, \cdots, p_L^0, u^0) \textrm{ for every } p_1>0 $$ $$ h(p_1^1, p_2, p_3^0, \cdots, p_L^0, u^1) = h(p_1^1, p_2, p_3^0, \cdots, p_L^0, u^0) \textrm{ for every } p_2>0 $$
Therefore, we have $$ EV(p^0, p^1, w) = CV(p^0, p^1, w) $$ if there is no wealth effect.