Choice-free one-dimensional ALE Reynolds transport identity

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 $\widehat{B}=\mathrm{\partial }\widehat{\mathrm{\Omega }}$. These boundaries are finite endpoint sets, hence compact. Covering each point by intervals of arbitrarily small total length shows that both have Jordan content zero and are compact Jordan sets. By [F10] the boundaries are null and the two open domains are Jordan measurable.

For any continuous $h$ on $K_t$, its zero extensions from $\mathrm{\Omega }(t)$ and $K_t$ to a nondegenerate closed interval containing $K_t$ in its interior are bounded and can be discontinuous only at $B_t$. By [F11] they are integrable. Their difference is supported on $B_t$ and bounded in absolute value by a constant times $1_{B_t}$; [F8] gives $\int 1_{B_t}=0$, so [F7] shows their integrals agree. The same argument applies on $\widehat{K}$ and $\widehat{\mathrm{\Omega }}$. The map image in [F3] is compact Jordan, and [F5] makes $g(\cdot ,t)$ and $G(\cdot ,t)$ integrable on the compact closures. Change of variables on those closures, followed by the open/closed equalities just proved, gives $I(t)={\int }_{\widehat{K}}G(\xi ,t)d\xi ={\int }_{\widehat{\mathrm{\Omega }}}G(\xi ,t)d\xi$, where $G(\xi ,t):=g(A_t(\xi ),t)J(\xi ,t)$.

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 latter and each quotient are integrable on $\widehat{K}$ by [F5]; [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, it follows that $I^{\mathrm{\prime }}(t)={\int }_{\widehat{K}}{\mathrm{\partial }}_{t}G(\xi ,t)d\xi$.

The chain and product rules together with [F2] give $I^{\mathrm{\prime }}(t)={\int }_{\widehat{K}}J({\mathrm{\partial }}_{t}g+{\mathrm{\partial }}_{x}g{w}_m+g{\mathrm{\partial }}_x{w}_m)(A_t(\xi ),t)d\xi$. This integrand is continuous on the compact reference closure; [F3] changes variables on the compact closures, and step 2.1 converts the resulting physical-domain integrals to open-domain integrals. Thus $I^{\mathrm{\prime }}(t)={\int }_{\mathrm{\Omega }(t)}[{\mathrm{\partial }}_{t}g+{\mathrm{\partial }}_x(g{w}_m)](x,t)dx$.

The function $x\mapsto g(x,t)w_m(x,t)$ is differentiable on each compact image interval and its derivative is continuous, hence integrable there by [F5]. Apply [F4] to obtain ${\int }_{{\alpha }_j}^{{\beta }_j}{\mathrm{\partial }}_x(g{w}_m)dx=g({\beta }_j,t)w_m({\beta }_j,t)-g({\alpha }_j,t)w_m({\alpha }_j,t)$. Summing right endpoints with normal $+1$ and left endpoints with normal $-1$ proves the boundary term. No choice axiom is used.