Integral over all internal points Let $X$ be some subset of real numbers, and $f$ a real-valued function on $X$. 
If  $ f(x) \geq 0$ for every point $x\in {X}$, then also:
$$\int_{x\in X} f(x)dx \geq 0.$$
What if $ f(x) \geq 0$ only for every point $x\in \text{interior}({X})$ - is it still true that 
$$\int_{x\in X} f(x)dx \geq 0 ?$$
I thought to claim that this is true because "almost all" points in $X$ are internal points (the boundary points have a measure of 0). It this true?
 A: No, because the boundary need not have measure zero. In fact $X$ could be a set of positive measure with empty interior.
For example, say $r_1,\dots$ are the rationals in $[0,1]$, $\delta_n>0$ and $\sum \delta_n<1/2$. Let $$X=[0,1]\setminus\bigcup_n(r_n-\delta_n,r_n+\delta_n).$$
Edit The OP says this is interesting. That's correct. Actually $X$ is a compact set of positive measure that contains no rationals. This seems interesting to me - in order to have positive measure $X$ must contain a lot of irrationals; one might think that with so many irrationals in there there would be a sequence converging to some rational,  but no.
The OP asks what about $I$, the set of irrationals in $[0,1]$?
Yes, $I$ is also a set of positive measure with empty interior. But $X$ is a... can't put my finger on the right adjective, "better" will have to do; $X$ is a better example, for the following reason:
Let's say $A\sim B$ if $A$ and $B$ differ by null sets (that is, $m(A\triangle B)=0$, or $B=(A\cup N_1)\setminus N_2$ where $N_1$ and $N_2$ are null sets).
When we're doing measure theory we tend to regard things that differ only on a null set as the same. Yes, $I$ is a set of positive measure with empty interior, but $I\sim[0,1]$; the fact that $I$ has empty interior is sort of cheating, not ignoring null sets. On the other hand, if $X'\sim X$ then $X'$ also has empty interior; $X$ "really" has empty interior, can't change that by adjusting things on a null set.
