Let $C(X)$ be the collection of all continuous real-valued functions defined on a compact metric space $X \subset \mathbb{R}$.
Let $C(X)_+$ be the positive cone of $C(X)$ (i.e., $C(X)_+ := \{ f \in C(X) \colon f \geq \theta \}$, where $\theta(x)\equiv 0$ for all $x \in X$ is the zero vector in $C(X)$.).
Let $\nu$ be a probability measure on the measurable space $(X, \mathcal{B}(X))$, where $\mathcal{B}(X)$ denotes its Borel $\sigma$-algebra of $X$.
It is well known from the elementary property of integral that $$ \int_X f(x) \nu (\mathrm{d} x) \geq 0$$ whenever $f \geq \theta$.
Furthermore, $ \int_X f(x) \nu (\mathrm{d} x) > 0$ holds true if $f$ is strictly positive (i.e., $f(x) >0$ for all $x \in X$).
I am wondering that what if $f$ is just non-zero element of the positive cone $C(X)_+$ (i.e., $f \in C(X)_+ \setminus \{\theta\}$), can we still conclude that its integral is strictly positive $$ \int_X f(x) \nu (\mathrm{d} x) > 0 \qquad ?$$
I guess that the answer is yes, since $f$ is continuous rather than just Borel measurable. But I have been struggling with this question for a while, so could any kind person help me out please?
Thank you very much in advance! Any idea or suggestions are much appreciated!
First of all there is no such thing as normal distribution on a compact metric space. Normal/Gaussian distribution id defined on Euclidean spaces or topological vector spaces but not on metric spaces. Your conclusion is true for Borel measures $nu $ on $X$ such that $\nu (U)>0$ for every non-empty open set $U$. A counter example to your statement is obtained by taking $\nu $ to be a delta measure at a point.