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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!

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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.

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  • $\begingroup$ Thanks so much for your kind advice, @Kavi Rama Murthy . I learned it. But I am still curious how to prove my statement above for a general probability measure? (I've changed my description a bit) Since I am quite keen on understanding this point and I am just a beginner in this area. Would you mind to provide me more details in explanation please? I really appreciate :-) $\endgroup$ – Paradiesvogel Jul 21 '18 at 23:45
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    $\begingroup$ I have given a counter example already. Suppose $\nu (A)=1$ if $x \in A$, $0$ otherwise, where $x \in X$ is fixed. Then $\int f \, d\nu =f(x)$. If $f$ is non-negative and not identically $0$ we cannot say that $f(x) >0$ for any $x$, right? So what you are trying to prove is simply not true. $\endgroup$ – Kabo Murphy Jul 22 '18 at 0:28
  • $\begingroup$ Thanks for your reply @Kavi Rama Murthy . It's a huge help to me. I understand the counterexample now. But I still confused that why the statement holds true if $\nu$ is normal distribution? Generally speaking, under which kind of probability measures, such integral (expectations) will be strictly positive? Sorry for bothering you, but I am sincerely keen on getting this point. Could you explain this please? Thanks again :-) $\endgroup$ – Paradiesvogel Jul 22 '18 at 0:36
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    $\begingroup$ I have answered this point also: if $\nu (U)>0$ for every non-empty open set $U$ the the conclusion is true. I particular if $\nu $ has a strictly positive density f (where $\nu $ is a Borel measure on $\mathbb R$) then the conclusion holds. In particular it is true for normal distribution. It is not true for exponential distribution (whose density is $0$ on $(-\infty ,0)$. $\endgroup$ – Kabo Murphy Jul 22 '18 at 11:38
  • $\begingroup$ Thanks for your kind explanation @kavi rama murthy . It’s very helpful to me and I learned from it. Thank you so much again:-) $\endgroup$ – Paradiesvogel Jul 22 '18 at 11:43

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