Show that D={t∈R:P(x∈E: f(x)=t)>0}D=\{t\in \mathbb{R}:P\left(x\in E:~f(x)=t\right)>0\} is at most countable.

Let (E,E,ρ)(E,\mathcal{E},\rho) be an arbitrary metric space, with E\mathcal{E} being the smallest σ\sigma-algebra containing the open subsets of EE induced by the metric ρ\rho. Let f:E→Rf:E\rightarrow \mathbb{R} be a continuous and bounded real-valued function defined on EE. Suppose PP is a probability measure defined on the measurable space (E,E)(E,\mathcal{E}). Show that the set

D={t∈R:P(x∈E: f(x)=t)>0}\begin{align}
D = \{t\in \mathbb{R}:P\left(x\in E:~f(x)=t\right)>0\}

is at most countable.

Any hints on solving this problem will be of help.



2 Answers


Hint: Since PP is a probability measure, the set

Dk:={t∈R;P(x∈E;f(x)=t)≥1k}D_{k} := \left\{t \in \mathbb{R}; P(x \in E; f(x)=t) \geq \frac{1}{k} \right\} is finite. Thus, D=⋃k∈NDkD = \bigcup_{k \in \mathbb{N}} D_k is at most countable.



Thank you so much.
– AGuyWhoWantsToSucceed
2 days ago



@AGuyWhoWantsToSucceed You are welcome.
– saz
2 days ago

For each t∈Rt\in \mathbb{R}, let
E_t = \{x\in E : f(x) = t\} = f^{-1}(\{t\})

For n∈Nn\in \mathbb{N}, let
D_n = \{t \in \mathbb{R} : P(E_t) > 1/n\}

Then it is clear that D=∞⋃n=1DnD = \bigcup_{n=1}^{\infty} D_n and so it suffices to show that each DnD_n is a finite set. In fact, we claim that
1 = P(E) \geq P(\sqcup_{i=1}^{n+1} E_{t_i}) = \sum_{t=1}^{n+1} P(E_{t_i}) > \frac{n+1}{n}

which is a contradiction. This proves the claim and with it, the result you seek.



Thank you so much for the detailed explanation.
– AGuyWhoWantsToSucceed
2 days ago