Example of a measure on $\mathcal{P}(X)$ Is there an example of a set $X$ and a measure $\mu$ on $\mathcal{P}(X)$ (collection of all the the subsets of $X$) such that $\mu(X)=1$ and $\mu(\{x\})=0$ for all $x$ in $X$?
At first I thought it's obvious that there should be such a measure but then I couldn't find any. Now I have a feeling it has something to do with being countably additive. but I don't know why. 
Is there any lemma or theorem that shows there can't be any measure with this property?
 A: At least under some set theoretic assumptions, such set and measure exist. Moreover, the set can be $\Bbb R$ and the measure can be an extension of the Lebesgue measure.
We say that a cardinal $\kappa$ is measurable if there is a non-principal ultrafilter $U$ on $\kappa$ which is $\kappa$-additive.


*

*The existence of a measurable cardinal is quite stronger than "standard set theory", i.e. $\sf ZFC$, in the sense that from the existence of a measurable cardinal we can prove that $\sf ZFC$ is consistent, and much much more. This means that we cannot prove that measurable cardinals are consistent, unless we start with even stronger hypotheses, but nevertheless the consensus seems to be that measurable cardinals are "probably consistent", as they have been studied extensively for several decades now.

*If $\kappa$ is the least uncountable cardinal such that $\kappa$ carries a $\sigma$-complete non-principal ultrafilter, then $\kappa$ is in fact a measurable cardinal. This is a non-trivial theorem, but it shows that if you want to strengthen your requirements that the measure is binary, then we already shoot quite high our the set theoretic assumptions.

*If we assume that $\kappa$ is in fact a measurable cardinal, then we can prove that there is a mathematical universe where $2^{\aleph_0}=\kappa$, and there is a measure $\mu$ on $[0,1]$ which measures all subsets and extends the Lebesgue measure. As per the axiom of choice, this $\mu$ is no longer translation invariant, but nevertheless it is $\sigma$-additive.
While I don't know what happens if allow both the measure to be non-binary, and the set to be larger than the reals. But if I were a gambling man, I would guess this is probably as strong a requirement as having a measure cardinal, in terms of consistency assumptions.

If you are willing to let go of the axiom of choice, then it is also possible that the Lebesgue measure on $[0,1]$ satisfies these requirements. However, one can see that in order to ensure that the measure is non-zero, one has to ensure that the countable union of countable sets of reals are countable. So some choice is needed, but it is consistent with $\sf DC$, which is a mild form of choice, that the Lebesgue measure in fact measures all the subsets of $[0,1]$.
It should be remarked that this again requires assumptions "going beyond $\sf ZFC$", although significantly weaker than those discussed above. Here a single inaccessible cardinal would suffice.
(If one is willing to allow an extension of the Lebesgue measure, and allows choice to fail, then indeed one can obtain this without any additional consistency assumptions beyond the standard set theory.)
A: Such measure does not exist.

$Theorem:$ Let $μ$ be a nonnegative bounded measure defined on all subsets of a set $S$.Then exists a countable set$C \subseteq S$ such that $μ(S$\ $C)=0$

Now let $X$ be a set and  suppose that exists a measure $μ$  on  $P(X)$ such that $μ(Χ)=1$ and $μ(\{x\})=0, \forall x \in X$.
Then exist a countable set $C \subseteq X$ such that $μ(X$\ $C)=0$
But $C=\{x_1,x_2....\}$ and $μ(C)=\sum_{n=1}^\infty μ(\{x_n\})=0$ thus $μ(X)=μ(Χ$\ $C)+μ(C)=0$ which is a contradiction.
I will also post down to the comments the text ,in which i found this theorem.
