The image of a lower-dimensional $C^1$ manifold is null

Under the stated Countable Choice, choose countable smooth atlases $\{(U_i,{\phi }_i)\}$ on $P$ and $\{(V_j,{\psi }_j)\}$ on $N$ as in [L3]. The countable family of open sets $U_i\cap F^{-1}(V_j)$ covers $P$. It is enough to show that the image of each is null in $N$, then use [L4] for their union.

Fix one open overlap $U=U_i\cap F^{-1}(V_j)$, write $\mathrm{\Omega }={\phi }_i(U)\subseteq {\mathbb{R}}^m$, and let $(V_j,{\psi }_j)$ be its target chart. Define $\tilde{F}:\mathrm{\Omega }\times {\mathbb{R}}^{n-m}\to {\mathbb{R}}^n,\tilde{F}(u,z):=({\psi }_j\circ F\circ {\phi }_i^{-1})(u)\mathrm{.}$ The slice $\mathrm{\Omega }\times \{0\}$ is Euclidean null: cover it by countably many bounded $m$-cubes times $\{0\}$; for each cube, finite $(n)$-cube covers of arbitrarily small total volume are obtained by choosing a sufficiently thin grid in the remaining $n-m\ge 1$ coordinates, and [F1] combines these with geometric error budgets under Countable Choice. The map $\tilde{F}$ is $C^1$. Thus [L1] makes $\tilde{F}(\mathrm{\Omega }\times \{0\})={\psi }_j(F(U))$ Euclidean null.

To check every target chart $V_k$, apply the same argument to the further open overlap $U\cap F^{-1}(V_k)$; its coordinate map has target $V_k$, so step 2.1 makes ${\psi }_k(F(U)\cap V_k)$ Euclidean null. The countable chart cover on $N$ detects $F(U)$ as null by [L2]. Finally [L4] makes the countable union $F(P)$ null.