$\newcommand{\pl}{\partial}$ $\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$
Let $\M,\N$ be smooth compact, connected, oriented manifolds of the same dimension.
Let $\,f:\M \to \N$ be a smooth immersion.
Does there exist a subset $A \subseteq \M$ of measure zero such that $f|_{\M \setminus A}$ is injective?
Edit: As shown be the answer of Lee Mosher, the interesting case is when $\partial M \neq \emptyset$. For instance, $f:[0,2\pi] \to \mathbb{S}^1$, $f(t)=e^{it}$. (Then $A=\{ 0,2\pi\}$).
In the case where $\partial M \neq \emptyset$, $f$ is not an open map, so Lee's argument that $f$ needs to be injective does not stand as is.
(Updated to take into account $\partial M \ne \emptyset$)
I'll write my answer using the Lebesgue measure class of a smooth volume form on $M$. The answer is that $A$ exists if and only if the restriction of $f$ to the manifold interior $\text{int}(M)$ is already injective, in which case you can take $A = \partial M$.
To prove this, suppose there exists $p \in \mathcal{N}$ and $x_1 \ne x_2 \in f^{-1}(p) \cap \text{int}(M)$. Then we may choose neighborhoods $U_1$ of $x_1$, $U_2$ of $x_2$, and $V$ of $p$, such that $f$ restricts to diffeomorphisms $U_1 \mapsto V$ and $U_2 \mapsto V$. Hence removal of a measure zero subset of $M$ leaves full measure subsets $A_1$ of $U_1$ and $A_2$ or $U_2$, each mapping to a full measure subset of $V$. The map $f$ is two-to-one over the full measure subset $f(A_1) \cap f(A_2)$ of $V$.