Reynolds transport identity on an ALE moving domain

Put $\widehat{K}=\overline{\widehat{\mathrm{\Omega }}}$, $K_t=A_t(\widehat{K})=\overline{\mathrm{\Omega }(t)}$, and let $B_t=\mathrm{\partial }\mathrm{\Omega }(t)$ and $\widehat{B}=\mathrm{\partial }\widehat{\mathrm{\Omega }}$. Cover each compact face in either finite boundary presentation by finitely many smaller regular $C^1$ graph charts with bounded base and bounded graph derivative. For a base grid of side $\delta$, each chart needs at most $C{\delta }^{-(d-1)}$ base cubes; the Lipschitz bound places each graph portion in an axis-parallel $d$-cube of side $C^{\mathrm{\prime }}\delta$. The total covering volume is at most $C^{\mathrm{\prime }\mathrm{\prime }}\delta$, which tends to zero. Since there are finitely many faces, both boundaries are compact Jordan sets of content zero. By [F11] they are null and the open domains are Jordan measurable; by the compact-image conclusion in [F4], $K_t$ is compact Jordan.

For any continuous $h$ on $K_t$, its zero extensions from $\mathrm{\Omega }(t)$ and $K_t$ to a nondegenerate closed rectangle containing $K_t$ in its interior are bounded and have discontinuities only on $B_t$, a null set, so [F12] makes them integrable. Their difference is supported on $B_t$ and bounded in absolute value by a constant times $1_{B_t}$. By [F8], $\int 1_{B_t}=0$, and [F7] shows the two integrals agree. The same argument applies to $\widehat{K}$ and $\widehat{\mathrm{\Omega }}$. By [F10], continuous integrands on the compact closures are integrable, so [F4] applies to the compact closures. Consequently, for $G(\xi ,t):=g(A_t(\xi ),t)J(\xi ,t)$, $I(t):={\int }_{\mathrm{\Omega }(t)}gdx={\int }_{\widehat{K}}G(\xi ,t)d\xi ={\int }_{\widehat{\mathrm{\Omega }}}G(\xi ,t)d\xi$.

For small nonzero $h$, apply [F6] to $s\mapsto G(\xi ,s)$ between $t$ and $t+h$. The difference quotient equals ${\mathrm{\partial }}_{t}G(\xi ,{\theta }_{\xi })$ at an intermediate time. By [F9], ${\mathrm{\partial }}_{t}G$ is uniformly continuous on $\widehat{K}$ times a compact time neighborhood of $t$, so these quotients converge uniformly in $\xi$ to ${\mathrm{\partial }}_{t}G(\xi ,t)$. The derivative and difference quotients are continuous on the compact Jordan set $\widehat{K}$, hence integrable by [F10]. [F7] and [F8] bound the integral of their difference by its uniform norm times $\mathrm{cont}(\widehat{K})$. Since $I(t)={\int }_{\widehat{K}}G(\xi ,t)d\xi$ by step 2.1, $I^{\mathrm{\prime }}(t)={\int }_{\widehat{K}}{\mathrm{\partial }}_{t}G(\xi ,t)d\xi$. The chain and product rules then give $I^{\mathrm{\prime }}(t)={\int }_{\widehat{K}}J({\mathrm{\partial }}_{t}g+{\mathrm{\nabla }}_{x}g\cdot w_m)+g{\mathrm{\partial }}_{t}Jd\xi$.

Substitute [F2] into step 3.1 and use ${\mathrm{\nabla }}_{x}g\cdot w_m+g{\mathrm{div}}_x{w}_m={\mathrm{div}}_x(g{w}_m)$. The resulting integrand is continuous on the moving closure. Apply [F4] to the compact closures and use the open/closed integral equality from step 2.1 to obtain $I^{\mathrm{\prime }}(t)={\int }_{\mathrm{\Omega }(t)}({\mathrm{\partial }}_{t}g+{\mathrm{div}}_x(g{w}_m))dx$.

Set $f={\mathrm{div}}_x(g{w}_m)$ and choose a nondegenerate closed box $Q$ containing $K_t$ in its interior. The field $g{w}_m$ is $C^1$ on the moving closure, so $f$ is continuous and bounded there. Its zero extension $\tilde{f}$ from the open set $\mathrm{\Omega }(t)$ to $Q$ is Borel: each strict superlevel is a relatively open subset of $\mathrm{\Omega }(t)$, together with $Q\setminus \mathrm{\Omega }(t)$ when the level is negative. Step 2.1 gives its Riemann integrability. By [A1] and [F13], its Riemann integral over $Q$ equals its Lebesgue integral, which is the Lebesgue integral of $f$ over $\mathrm{\Omega }(t)$. Thus the Jordan volume integral in step 4.1 is exactly the volume integral used by [F5]. Apply [F5] to $g{w}_m$ on the finite piecewise-$C^1$ presentation [F3] to obtain ${\int }_{\mathrm{\Omega }(t)}{\mathrm{div}}_x(g{w}_m)dx={\int }_{\mathrm{\partial }\mathrm{\Omega }(t)}g{w}_m\cdot ndS$. Substitution into step 4.1 proves the asserted identity. Countable Choice is used through [F5] and [F13].