Question about the definition of concentration function $\alpha_{P, X, \rho}(\epsilon):=\sup_{A\subset X}\{1-P(A^{\epsilon})|P(A)\ge 1/2\}$.

Question

I have a question about the definition of concentration function, which is defined as in Wainwright, High-dimensional statistics.

The concentration function $\alpha:[0,\infty)\to \mathbb{R}_{+}$ associated with metric measure space $(P, X, \rho)$ is given by $$ \alpha_{P, X, \rho}(\epsilon):=\sup_{A\subset X}\{1-P(A^{\epsilon})|P(A)\ge 1/2\} $$ where the $A^{\epsilon}:=\{x\in X|\rho(x,A)<\epsilon\}$ and the supremum is taken over all measurable subsets $A$.

Q1: What does "$\sup_{A\subset X}\{1-P(A^{\epsilon})|P(A)\ge 1/2\}$" mean? I am confused about the notation $|$ inside the sup. Is that "$\sup E(\cdot|\cdot)$"?

Q2: Why we have $\alpha_{P, X, \rho}(\epsilon)\ge 1/2$?

Answer 1

The vertical line | means "such that". That is, the concentration function measures the supremum of $1 - P(A^\epsilon)$ over choices of $A \subset X$ with $P(A)\geq 1/2$. Another way of writing this is:

$$\alpha_{P,X,\rho}(\epsilon) = \sup_{ \substack{A \subset X, \\ P(A)\geq 1/2}}(1-P(A^\epsilon))$$

We don't have $\alpha_{P,X,\rho}(\epsilon) \geq 1/2$. Instead, note that by monotonicity, $P(A^\epsilon) \geq P(A)$, so that $$1-P(A^\epsilon) \leq 1-P(A) \leq \frac{1}{2}$$ for every event $A$ with $P(A) \geq 1/2$. Taking the supremum, we instead get $$\alpha_{P,X,\rho}(\epsilon) \leq 1/2$$