Path Integral Measure Transformation as $(DetU)^{-1}$

Question

The path integral measure transforms as $D\Psi\rightarrow (DetU)^{-1}D\Psi$ for fermions, with $DetU=J$ the Jacobian.
I am referring to Peskin and Schroeder's Introduction to Quantum Field Theory, Chapter 19 about chiral anomalies. What is the intuition behind the path integral measure transforming by the inverse of the determinant for fermions? Is this just by definition/convemtion? I am struggling to see the significance of this choice.

Answer 1

Grassmannian integrals satisfy (this is a definition) $$ \int d \theta \theta = 1 \tag{1} $$ Let $\theta' = A \theta$. Then, $$ 1 = \int d \theta' \theta' = \int d (A\theta) (A\theta) = A \int d (A\theta) \theta $$ Consistency with (1) then requires $$ d(A\theta) = \frac{1}{A} d \theta. $$

On the other hand, bosonic integrals satisfy $$ \int dx x \sim x^2 \tag{2} $$ Let $x'=Ax$. Then, $$ x'^2 \sim \int dx' x' \implies (Ax)^2 \sim \int d(Ax) (Ax) \implies x^2 \sim \frac{1}{A} \int d(Ax) x $$ Consistency with (2) requires $$ d(Ax) = A dx. $$ The generalization to multiple variables $\theta \to \theta_i$, $x \to x_i$ and $A \to \det A$ can be easily checked.