Showing a result for simple functions

Solution 1:

You can prove this for simple functions $L=\sum_{i=1}^nc_i\mathbb 1_{A_i}$ by using linearity of conditional expectation as follows:

$ {\displaystyle \int } \sum_{i=1}^nc_i\mathbb 1_{A_i} d\delta_1(X,a) = {\displaystyle \sum_{i=1}^n }c_i \delta_1(X,A_i)$ by definition

$\hspace{44mm}={\displaystyle \sum_{i=1}^n c_i\mathbb E\left[\delta_0(X,A_i)\;|\; T \right]}$

$\hspace{44mm}=\mathbb E\left [{\displaystyle \sum_{i=1}^n c_i\delta_0(X,A_i)\;\Bigg|\; T } \right]$ by linearity of conditional expectation

$\hspace{44mm}=\mathbb E\left [{\displaystyle \int} \sum_{i=1}^n c_i\mathbb 1_{A_i} d\delta_0(X,a)\;\Bigg|\; T \right]$

$\hspace{44mm}=\mathbb E\left [{\displaystyle \int} L\; d\delta_0(X,a)\;\Bigg|\; T \right]$ as desired.