Finite measure on the power set Let $X$ be an uncountable set, and let $\Omega$ be the power set of $X$, viewed as a $\sigma$-algebra. Does there exist a positive $\sigma$-additive measure of finite total mass on $(X, \Omega)$ such that each point of $X$ has measure zero? 
 A: Just to complement Andres's excellent answer with another reference, you can find a nice summary of the status of this question, as well as further references, in chapter 1.12(x) of Bogachev's monograph "Measure Theory I". 
The (very) short summary is that in all concrete cases the answer is no.
A: I assume you mean a $\sigma$-additive measure. This is Ulam's measure problem. A positive answer is closely tied up to the existence of real-valued measurable cardinals, so it is equiconsistent with the existence of a measurable cardinal, which is a large cardinal assumption significantly beyond the usual axioms of set theory.
You can see a quick write up of the argument here. A good reference is the beginning of David Fremlin, "Real-valued measurable cardinals", in Set Theory of the reals, Haim Judah, ed., Israel Mathematical Conference Proceedings 6, Bar-Ilan University (1993), 151–304, that I also mention in the notes linked to above.
In short (this is expanded in the notes): If $(X,\mathcal P(X),\lambda)$ is such a measure space, we may as well assume (by concentrating on an appropriate subset, which may be of smaller size than $X$, and renormalizing) that $\lambda$ is a probability measure. Its additivity is the smallest cardinal $\kappa$ such that the measure of the disjoint union of some collection of $\kappa$ many disjoint subsets of $Y$ is not the sum of the measures of the sets in the union. (So the additivity is at least $\aleph_1$, and it is well-defined, since we are assuming that $\lambda(X)>0$.)
Then we can in fact assume $X=\kappa$ (identifying cardinals with sets of ordinals). If $\lambda$ is non-atomic (meaning, for any $E\subseteq\kappa$, if $\lambda(E)>0$ then there is $F\subset E$ with $0<\lambda(F)<\lambda(E)$), then $\lambda$ is (atomlessly) real valued measurable. On the one hand, these cardinals are not too large: $\kappa\le|\mathbb R|$. On the other, $\kappa$ must be weakly inaccessible, and in fact limit of weakly inaccessibles that themselves are limit of weakly inaccessibles, etc. This is very very large.
The other possibility is that $\lambda$ is atomic. Then, after further renormalization, $\lambda$ can be identified with the characteristic function of a non-principal $\kappa$-complete ultrafilter, that is, $\kappa$ is measurable. 
