We've defined a finite element as a triple $(T,P,K)$ where $T$ is the element, $P$ is a space of polynomials and $K=\{\chi_0,...,\chi_R\}$ is a set of node functionals $\chi_j:C^\infty(T) \rightarrow \mathbb{R}$ and there exists $m \geq 1$ with $P_{m-1} \subset P$ and there exists $p \in [1, \infty]$ such that every $\chi \in K$ extends to a bounded linear operator on $W^{m,p}(T)$.
Now we have the following example in the script: Let $T \subset \mathbb{R}^d$ be a closed subset with $d=1,2,3$. The triple $(T, P_0, \chi_T)$ with $\chi_T(v)= \int_T vdx$ is a finite element with $m=1$ and $p=1$. The claim that $m=1$ is clear, but I don't understand why we obtain $p=1$. This means that $\chi_T$ can be extended to a bounded linear operator on $W^{1,1}(T)$ but why is this true?
The functional $$ v\mapsto \int_T v \ dx $$ is a bounded linear functional on $L^1(T)$, as $$ \left |\int_T v \ dx\right|\le \int_T |v| \ dx=\|v\|_{L^1(T)}. $$ So you could choose $p=0$. Since $W^{1,1}(T)$ is continuously embedded into $L^1(T)$ (trivially), the functional is also continuous on $W^{1,1}(T)$. In fact, any choice of $p\ge0$ is valid here.