Integrating over a countable set Suppose $f(x)$ is strictly positive and continuous over an open interval $S=(a,b)$, so that $\int_{S} f(x) dx > 0$. We commonly encounter integration over these uncountably infinite sets like $S$ in evaluating integrals. My question is, is it ever possible to have a finite, or countably infinite, subset $T \subset S$, such that $\int_{T} f(x) dx > 0$? 
My initial thought process went like this. Suppose $U \subset S$ is a subset of only one element, $U = \{ c : c \in (a,b) \}$. Then integrating over $U$ yields $\int_{U} f(x) dx = \int_c^c f(x) dx = 0$. Extending this, it seems if we create the finite set $V = \{ x_k : x_k \in S, k = 1,...,n \}$, for example, some $n$ integers lying in $(a,b)$, then $\int_{V} f(x) dx = \sum_{k=1}^{n} \int_{x_k}^{x_k} f(x) dx = 0$. So, is there a set $T\subset S$ that circumvents this? Thanks. 
 A: You are indeed correct: any integral over a set of measure zero is equal to zero, even if it is uncountable.
The value of the integral has little to do with the number of elements that lie in the set you choose and more with "how they are spaced".
You would have to use a different measure, i.e., that $dx$ would have to be replaced by something else. You stumbled upon a very wide topic. Checkout some real analisys text, though I wouldn't be able to get you a reference myself. 
A: You're correct in thinking that, when using the usual Lebesgue integral and true functions $f$, the integral over a countable set must always be zero. Your argument only covers finite sets, though, not countably infinite ones. To show it for a countably infinite set $T$, recall that countably infinite sets have measure $0$, and that the integral over a set of measure zero is zero.
Now, that's not the end of the story. I and other commenters have pointed out that this is only true for the Lebesgue integral - if integrate with respect to a different measure, like the counting measure or a measure that happens to be concentrated on $T$, then the integral will often not be zero. There's also something called the Dirac $\delta$ function, which is what's called a generalized function, defined so that $\int_{\{0\}}\delta d\mu = 1$ - that is, even when integrating with respect to Lebesgue measure, the integral of $\delta$ over a single point can be nonzero. $\delta$ isn't really a function - it doesn't make sense to ask what $\delta(0)$ is, for example - but it's a useful fiction used in probability theory (among other things).
A: If we are talking here about the Lebesgue integral, the answer is: no, because finite and countable infinite subsets of $\mathbb R$ have Lebesgue measure equal to $0$.
If we are talking about the Riemann integral, the answer is the same, although the proof is somewhat different.
