Finite-box stress balance and the flux form of conservation

Claim (a). Fix $b$ and $a$. For each fixed tuple $y$ of the remaining three coordinates, the function $t\mapsto T^{ab}(t,y)$ is $C^1$ on $[c_a,d_a]$ as a function on that interval, with derivative ${\mathrm{\partial }}_a{T}^{ab}(t,y)$, which is continuous and hence integrable by [F4]; [F1] gives ${\int }_{c_a}^{d_a}{\mathrm{\partial }}_a{T}^{ab}(t,y)dt=T^{ab}(d_a,y)-T^{ab}(c_a,y)$. The function $(t,y)\mapsto {\mathrm{\partial }}_a{T}^{ab}(t,y)$ is continuous on the box $B$, so its repeated integral in this order exists and equals ${\int }_B{\mathrm{\partial }}_a{T}^{ab}{d}^{4}x$ by [F5] and [F2]; substituting the pointwise identity above gives ${\int }_B{\mathrm{\partial }}_a{T}^{ab}{d}^{4}x={\int }_{F_a^{+}}{T}^{ab}{d}^{3}x-{\int }_{F_a^{-}}{T}^{ab}{d}^{3}x$. The integrands are continuous on the box $B$, hence integrable by [F4]; summing the identities over $a=0,1,2,3$ with the linearity part of [F3] yields the box identity (a).

Claim (b), finiteness and continuity of $Q^b$. For each $s\in [c_0,d_0]$ the function $x\mapsto T^{0b}(s,x)$ is continuous on the compact box $K$, hence bounded and integrable by [F4] and [F6], so $Q^b(s)$ is a finite real number. For $s,r\in [c_0,d_0]$ and each $x\in K$, [F1] applied to $t\mapsto T^{0b}(t,x)$ gives $T^{0b}(s,x)-T^{0b}(r,x)={\int }_r^s{\mathrm{\partial }}_0{T}^{0b}(t,x)dt$; integrating over $K$ with [F2] and [F5] gives $Q^b(s)-Q^b(r)={\int }_K{\int }_r^s{\mathrm{\partial }}_0{T}^{0b}(t,x)dtdx$. By [F6] the continuous $\mathrm{\mid }{\mathrm{\partial }}_0{T}^{0b}\mathrm{\mid }$ is bounded on the compact box $B$ by some $M_1<\mathrm{\infty }$, so the absolute-value estimate [F3] and the constant integral give $\mathrm{\mid }{Q}^b(s)-Q^b(r)\mathrm{\mid }\le M_1\mathrm{vol}(K)\mathrm{\mid }s-r\mathrm{\mid }$, and $Q^b$ is (Lipschitz) continuous on $[c_0,d_0]$.

Claim (b), slice-balance identity. The sub-box $[s_1,s_2]\times K$ is a closed nondegenerate axis-parallel box contained in $B$, and $T$ is $C^1$ on a neighbourhood of $B$, hence on a neighbourhood of the sub-box. Applying the box identity of step 1.1 to this sub-box, the two faces with $a=0$ give ${\int }_K{T}^{0b}(s_2,x)d^{3}x-{\int }_K{T}^{0b}(s_1,x)d^{3}x=Q^b(s_2)-Q^b(s_1)$, and for $i=1,2,3$ the faces $F_i^{\pm }$ are the side faces $[s_1,s_2]\times K_i^{\pm }$ with the displayed signs; with the sub-box's interior equal to $(s_1,s_2)\times \mathrm{int}K$, this is exactly the stated identity.

Claim (c). If $f=0$ on $[s_1,s_2]\times K$ then its integral there vanishes, and if $T^{ib}=0$ on the side faces $[s_1,s_2]\times K_i^{\pm }$ then the three side-flux differences vanish; step 2.1 then gives $Q^b(s_2)-Q^b(s_1)=0$, so $Q^b(s_1)=Q^b(s_2)$, for every $b$.

Claims (a), (b) and (c) are proved in steps 1.1, 1.2 and 2.1, and 3.1. The argument uses only the fundamental theorem of calculus, Fubini's theorem for continuous functions on boxes and the elementary integral estimates; it concerns one inertial chart and finite axis-parallel boxes, asserts nothing for arbitrary regions, hypersurfaces or noncompact slices, and asserts no conservation without the stated side-flux hypothesis. All choices are finite; no choice principle is used.