I am working on a joke real analysis course with some friends and I have a problem which I managed to reduce to the existence of a function $P:2^{\mathbb{R}}\to\mathbb{R}$ for a fixed $a\in\mathbb{R}$ which satisfies the following:
1)$P(\mathbb{R})=1$
2)$\forall A \in 2^{\mathbb{R}} :P(A)\ge 0$
3)$P(A)+P(B)=P(A\cup B)+P(A\cap B)$
4)$\forall A \in 2^{\mathbb{R}}: a\notin \overline{A-\{a\}} \implies P(A)=0$
So a probability measure with only a finite sum property but the fixed accumulation point property added.I suspect such a function is impossible because of an argument such as vitali sets for the lebesgue measure, but i do not have the expertise to confirm this.
If it is possible to construct such a function, an additional question, which makes it more of a set theory question, is whether such a function exists with image {0,1}.
In that case the criteria can be rewritten as the existence of a set $H\subset 2^{\mathbb{R}}$ which satisfies the following:
1)$\mathbb{R}\in H$
2)$\forall A,B \in 2^{\mathbb{R}}: A\subset B \land A\in H \implies B\in H$
3)$\forall A,B \in 2^{\mathbb{R}}:$
-a)$A\in H \land B\in H \implies A\cap B \in H$
-b)$A\cup B \in H \implies A\in H \lor B\in H$
4)$\forall A \in 2^{\mathbb{R}}: a\notin \overline{A-\{a\}} \implies A\notin H$
My question then is, does such a function exist? And if it does, does a set such as described beneath it exist?