Consider the $n$-dimensional autonomous system $\frac{d}{dt}\boldsymbol{y}(t)=f(\boldsymbol{y}(t))$ defined on $[a,b]^n$, where $\mbox{-}\infty<a<b<\infty$.$^1$
Suppose that $\tilde{\boldsymbol{y}}$ is a steady state (i.e. $f(\tilde{\boldsymbol{y}})=0)$ but is not Lyapunov stable.$^2$ Does it then follow that its basin of attraction
$$\{\boldsymbol{y}(0)\in[a,b]^n: \lim_{t\to\infty} \boldsymbol{y}(t)= \tilde{\boldsymbol{y}} \}$$ has Lebesgue measure zero?
Footnotes:
$\quad $ 1. By this I mean that $f:[a,b]^n\to\mathbb{R}^n$ is defined in a way such that $\{\boldsymbol{y}(t):t\geq 0\}\subseteq [a,b]^n$ whenever $\boldsymbol{y}(0)\in[a,b]^n$. (And initial conditions not in $[a,b]^n$ are excluded from analysis).
$\quad 2.$ In the present context, $\tilde{\boldsymbol{y}}$ is Lyapunov stable if $\forall \varepsilon>0$, $\exists \delta>0$ s.t. if $\boldsymbol{y}(0)\in[a,b]^n$ and $||\boldsymbol{y}(0)-\tilde{\boldsymbol{y}}||_2<\delta$, then $||\boldsymbol{y}(t)-\tilde{\boldsymbol{y}}||_2<\varepsilon$ $\forall t\geq 0$, where $|| \cdot ||_2$ denotes the Euclidean norm.