The null-set definition is independent of the smooth atlas

Assume $E$ is $\mathcal{A}$-null and fix $(V,\psi )\in \mathcal{B}$. The overlaps $V\cap U$ with $(U,\phi )\in \mathcal{A}$ cover $V$. At each point of each overlap, [L1] gives a smaller open neighbourhood $W$ on which the coordinate transition $\psi \circ {\phi }^{-1}$ preserves Euclidean nullity in both directions. These $W$ form an open cover of $V$. Apply [L2] under the given countable choice to obtain a countable cover $(W_j)$ subordinate to these local transition neighbourhoods, and use [A1] to fix a containing local transition neighbourhood and its source chart $(U_j,{\phi }_j)$ for each $j$.

Since $E$ is $\mathcal{A}$-null, each ${\phi }_j(E\cap W_j)$ is null as a subset of ${\phi }_j(E\cap U_j)$. The chartwise conclusion of [L1], on the local neighbourhood containing $W_j$, gives that $C_j=\psi (E\cap W_j)$ is Euclidean null. The countable family $(W_j)$ covers $V$, so $\psi (E\cap V)={\bigcup }_j{C}_j$.

Fix $\epsilon >0$. Each null $C_j$ admits a sequence of closed-cube covers with total volume at most $\epsilon 2^{-j-1}$. Use [A1] to choose these sequences simultaneously. Flatten them by [L3]; the resulting sequence covers ${\bigcup }_j{C}_j$ and has total volume at most ${\sum }_j\epsilon 2^{-j-1}=\epsilon$. Thus $\psi (E\cap V)$ is null. Since $(V,\psi )$ was arbitrary, $E$ is $\mathcal{B}$-null. Interchanging $\mathcal{A}$ and $\mathcal{B}$ gives the reverse implication.