A function $f:\mathbb{R}_+\to \mathbb{R}^n$ is called piecewise-$\mathcal{C}^k$ if
- There is a partition $\{ x_i \}_{i=0}^\infty $ with $x_0=0$ and $x_{i+1}>x_i$ such that $f$ is $\mathcal{C}^k$ on each $\left( x_i,x_{i+1} \right)$.
Or
- $f$ is a $\mathcal{C}^k$ function except at a subset of $\mathbb{R}_+$ with zero Lebesgue measure.
Are these definitions correct and also, equivalent? (Honestly, I don't have a solid background on Lebesgue measure theory so I'm not sure about the second definition)
They're not correct and also not equivalent. (1) is close. In fact $f$ is piecewise $C^k$ if there is a partition such that $f$ is $C^k$ on each $(x_j,x_{j+1})$, and the $k$-th derivative $f^{(k)}$ has one-sided limits at each endpoint (which implies that $f^{(j)}$ also has one-sided limits for $0\le j < k$).
If we were talking about a bounded interval $(a,b)$ that would be a finite partition. For $(0,\infty)$ it could also be an infinite partition as long as $x_j<x_{j+1}$ and $x_j\to\infty$.
On the other hand, it's not clear to me what (2) even means! Say $m(E)=0$; what does it mean to say $f$ is $C^k$ on $\Bbb R\setminus E$? If $E$ is closed this is clear...