Proposition: Density weighting retains a derivative term

For $\sigma \in C^{\infty }(M)$ and a compact smooth vector field $J$,

When $\mathrm{d}\sigma$ is nonzero, the right side need not vanish. Thus an unweighted total-divergence equivalence does not imply equality after density weighting.

Proof. The integral of ${\partial }_{\mu }(\sigma J^{\mu })$ vanishes by compact support. The product rule gives the identity.

If ${\partial }_{\nu }\sigma$ is nonzero on an open set, choose a nonnegative nonzero compact test $\chi$ there. Set $J^{\nu }=\chi {\partial }_{\nu }\sigma$ and other components zero. The residual equals $-\int \chi ({\partial }_{\nu }\sigma {)}^2{\mathrm{d}}^{4}x\ne 0$.