Let M , N be smooth manifolds, and let F : M → N be...

No: a covering map of smooth manifolds is an immersion but (I'm nearly sure) never has this property unless it is a diffeomorphism.
For example, map $S^1=\{z\in \mathbb{C}\mid |z|=1\}$ to itself by $z\mapsto z^2$. This is an immersion but a measurable subset $T$ of the domain on which the map is injective has measure at most $\frac{1}{2}$ the measure of $S^1$, because if $A:S^1\to S^1$ is the antipode ($A(z)=-z$) then $T\cap A(T)=\mathrm{\varnothing }$, so that $\lambda (T)+\lambda (A(T))\le \lambda (S^1)$, where $\lambda$ is the length measure on $S^1$; but $\lambda (T)=\lambda (A(T))$, since $A$ is an isometry.