The above is a theorem from a text on nonlinear analysis. I have a question on the highlighted yellow part. We know the inverse function theorem gives us the existence of open neighbourhood such that $f$ is a diffeomorphism , but the inverse function theorem does not tell us what the sign of the determinant of $f$ is on this neighbourhood, so I want to ask why can we shrink the neighbourhood such that the sign of the determinant is constant?
Also for a related question, for a continuously differentiable mapping, is the integral $\int_\Omega \text{det} Df(x) $ a well defined integral in the sense that it gives us a finite value? Here $\text{det} Df(x) $ denotes the determinant of the Jacobian matrix
Both of your questions are answer by the continuity of the determinant. For the first one, suppose WLOG that $J_f(x_i)>0$ then by continuity of the determinant there exist a nhd $V$ of $x_i$ s.t. $J_f(x)>0$ on $U$. This is what the author meant by "shrinking the nhd further if necessary", he means that the IFT guarantees there exist a nhd $U_i$ of $x_i$ s.t. $f$ is a homeomorphism on $U_i$, then intersect this nhd with $V$ if necessary to preserve the sign of the determinant.
For the second one the answer is yes. Since $f$ is continuously differentiable on $\bar{\Omega}$, then all its partial derivatives are continuous, the determinant is a sum of products of partial derivatives hence is a sum of products of continuous functions and hence is continuous and thus integrable.