Here is the problem and my work toward a proof:
Question: Prove that in the following definition, the value of $\int_E f dx$ is independent of the choice of rectangle $J$:
Definition: $f:E\to\mathbb{R}$, bounded, is integrable on $E \subset \mathbb{R}^2$, with $E$ bounded, if there exists a rectangle, $J$, with $E \subset J$ and the auxiliary function
$$ \hat{f}_J(x):=\left\{ \begin{array}{lr} f(x) & x \in E\\ 0 & x \notin E \end{array} \right. $$
is an integrable function on the rectangle $J$. In this case, we define
$$ \int_E fdx:= \int_J \hat{f}_J dx $$
Proof: I suppose there are two rectangles, $J_1$ and $J_2$ so that $E \subset J_1$ and $E \subset J_2$. Then if $\hat{f}_{J_1}$ is integrable on $J_1$ and $\hat{f}_{J_2}$ is integrable on $J_2$, then the upper and lower integrals $I_{up}(\hat{f}_{J_1},J_1)=I_{low}(\hat{f}_{J_1}$ and $I_{up}(\hat{f}_{J_2},J_2)=I_{low}(\hat{f}_{J_2},J_2)$. This means that the infimums of the upper sums equal the supremums of the lower sums...
And I'm not seeing where to go from here. Of course I want to show that the value of the integral is equal for both choices of rectangle -- that is the goal. Thank you very much for your help!!