I am reading the "Foundation of machine learning" by Mehryar Mohri (https://cs.nyu.edu/~mohri/mlbook/). In the proof of Theorem 11.8, it said the following statement, which I can not understand.
Let $H$ be a family of real-valued functions, and\begin{align}G=\{ (x,y)\rightarrow L(h(x),y):h\in H\}.\end{align}Then, we define the another family of function\begin{align}C=\{c(h,t):(x,y)\rightarrow 1_{L(h(x),y)>t} ,h\in H, t\ge 0 \}.\end{align}
Then, the author said that the VC dimension of $C$ is equal to the pseudo dimension of $G$.
How can we prove this?
To show the VC dimension of $C$, we have to find such $\{(x_1,y_1), \ldots,(x_d,y_d) \}$ satisfying the following: for any binary vector $\mathbf{b}\in \{0,1\}^d $, there exists $(h,t)$ such that\begin{align}c(h,t)(x_i,y_i) = 1_{h(x_i)-y_i>t}=b_i, \forall i\end{align}Note that in the definition above, $t$ is equal for all $i$, but changes over the selection of $h$.
However, in the definition in (11.3) of the book, we know that the $Pdim(G)$ is equivalent to the modified VC dimension given by the following. There exist $\{(x_1,y_1,t_1), \ldots,(x_d,y_d, t_d) \}$, for any binary vector $\mathbf{b}\in \{0,1\}^d $, there exists $h$ such that\begin{align} 1_{h(x_i)-y_i>t_i}=b_i, \forall i\end{align}Note that in the definition above, the $t$ is different for all $i$, but fixed for the selction of $h$.