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.

  • $\begingroup$ 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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