Weak differentiation ignores null-set changes

Fix $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ and put $K=\mathrm{supp}\phi$. By [F4], $\mathrm{\mid }K\mathrm{\mid }<\mathrm{\infty }$. For an $L^p$ representative $f$, [F3] applied to $\mathrm{\mid }f\mathrm{\mid }$ and ${\mathbf{1}}_K$ gives ${\int }_K\mathrm{\mid }f\mathrm{\mid }<\mathrm{\infty }$: for $p<\mathrm{\infty }$ use $\mathrm{\parallel }{\mathbf{1}}_K{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}=\mathrm{\mid }K{\mathrm{\mid }}^{1\mathrm{/}{p}^{\mathrm{\prime }}}$ (with $p^{\mathrm{\prime }}=\mathrm{\infty }$ when $p=1$), and for $p=\mathrm{\infty }$ use ${\int }_K\mathrm{\mid }f\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }K\mathrm{\mid }$. The derivatives $D^{\alpha }\phi$ are bounded and supported in $K$, so all four test integrals in [F1] are finite.

Since $u=\tilde{u}$ almost everywhere, $u{D}^{\alpha }\phi$ and $\tilde{u}{D}^{\alpha }\phi$ are equal almost everywhere and integrable by step 1.1. By [F5] their integrals agree. The same argument applied to $v\phi$ and $\tilde{v}\phi$ gives equal right sides, including the sign $(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}$.

Therefore the weak identity for $(u,v)$ holds for this test exactly when the weak identity for $(\tilde{u},\tilde{v})$ does. The test was arbitrary, proving both directions. By [F2] this is precisely independence from the representatives of $L^p$ classes.