= Solution
Use the one-way density per unit length $g_+/L=2/(h|v|)$. The electron-flux magnitude in an interval is charge times velocity times line density:
$$
dI^+=e|v|\frac{2}{h|v|}f(\epsilon,\mu-eV)\,d\epsilon=\boxed{\frac{2e}{h}f(\epsilon,\mu-eV)\,d\epsilon}.
$$
If $g$ is instead defined for a finite length $L$, the traversal frequency is $|v|/L$; the same cancellation follows. The printed $I=Nev$ therefore requires $N$ to be a line density, not an unscaled total particle number.
Subtract the opposite reservoir's incoming flux. With the stated <electron> <electrochemical potentials>, conventional positive current is opposite to electron-number flux, so
$$
I=\frac{2e}{h}\sum_i\int_{E_i}^\infty\left[f(\epsilon,\mu)-f(\epsilon,\mu-eV)\right]d\epsilon.
$$
At low <temperature> and small bias, each fully transmitted occupied spin-degenerate subband contributes $2e^2/h$. Hence \b[$\boxed{G=N_b\,2e^2/h}$]. More generally, the <Landauer formula> weights each mode by its <transmission probability>; imperfect transmission removes exact integer plateaux.
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