Expected measure of a ball in a probability space with a metric Assume we are given a probability space $(\mathbb{X}, \mathcal{X}, \mathbb Q)$ and a measurable distance function defined on it $d:\mathbb{X}\times \mathbb{X}\to \mathbb{R}^+\cup\{0\}$ that conforms to the usual definition of distance on metric spaces. I am trying to understand (in lay terms) the &quot;expected value of the measure of a ball with fixed radius $r&gt; 0$ centered at a randomly chosen point in $\mathbb{X}$ with measure $\mathbb{Q}$&quot;. That is more formally
\begin{equation}
\int_{\mathbb{X} }q(x)\mathbb{Q}(\mathrm{d}x)  
\end{equation}
where $q(x) = \mathbb{Q}(B(x,r))$ and $B(x,r) = \{y\in \mathbb{X}: d(x,y)\leq r\} $. The first thing to ask here is if $q(x)$ is measurable but for now I assume $\mathcal{X}$ is large enough that it is.
My problem originates from this question: &quot;find a lower bound for the probability that a randomly chosen iid set of points in a space are close-by&quot;. I am wondering what assumptions are needed from $\mathbb{Q}$ and $d$ such that this can be lower bounded. Of course I would not expect a bound valid for all $\mathbb{Q}$ and $d$ but any tip about how to treat this integral (or you know if anybody thought about this before and gave it a name so I can search) would be helpful. The trivial bound $\inf_{x\in\mathbb X} \mathbb{Q}(B(x,r))$ is not useful for my purposes.
How is it defined the probability density function $\mathbb{Q}$? Is it discrete or continuous?
If it's continuous then $\lim_{r\rightarrow0}\int_{B_r}q(x)dx=0$
And if it's discrete then $\lim_{r\rightarrow0}\int_{B_r}q(x)dx=q(y)$
How is it defined the probability density function $\mathbb{Q}$? Is it discrete or continuous?
If it's continuous then $\lim_{r\rightarrow0}\int_{B_r}q(x)dx=0$
And if it's discrete then $\lim_{r\rightarrow0}\int_{B_r}q(x)dx=q(y)$