Sigma finite measure positive on uncountable subset of the reals We can construct a sigma finite (finite even!) measure $\mu$ which assigns a positive measure to every member of a countable subset $A\subset \mathbb R$ by writing $$A=\{q_1, q_2, ...\}$$ and defining $$\mu(B)=\sum_{q_n\in A}\frac{1}{2^n}$$
Can we construct a sigma-finite measure on $\mathcal P (\mathbb R)$ which is positive on every member of some uncountable subset of $\mathbb R$. Intuition tells me that we shouldn't be able to since this will create a "too big mass of measure" which won't allow sigma finitness. If we can't construct such a measure, how can we prove it is impossible?
 A: Not possible.
Suppose $A$ is uncountable and $\mu(\{x\}) > 0$ for every $x \in A$.  If $\mu$ is $\sigma$-finite then we can write $A = \bigcup_{n=1}^\infty A_n$ where $\mu(A_n) < \infty$.  Since $A$ is uncountable, one of the $A_n$ must be uncountable (otherwise $A$ would be a countable union of countable sets, which must be countable).  So let's say $A_1$ is uncountable.
Now for every $x \in A_1$, we have $\mu(\{x\}) > 0$, and therefore there is an integer $k$ (depending on $x$) such that $\mu(\{x\}) > 1/k$.  So if we let $B_k = \{x \in A_1 : \mu(\{x\}) > 1/k\}$, then $A_1 = \bigcup_{k=1}^\infty B_k$.  Then one of the $B_k$ must be infinite (else $A_1$ would be countable).  But since every element of $B_k$ has measure at least $1/k$, this implies $\mu(B_k) = \infty$ which is a contradiction.
A: It is indeed not possible. Let $U$ be an uncountable set and let $\mu$ be a measure such that $\mu\left(\{x\}\right)$ is positive for any $x$. Define $U_n:=\left\{x\in U, \mu\left(\left\{x\right\}\right) \gt 1/n\right\}$. Then $U=\bigcup_{n\in\mathbb N^*}U_n$, and since $U$ is uncountable, then one of the $U_n$'s is not countable, say $U_{n_0}$. 
We thus have a measure $\mu$ (the restriction is denoted in the same way for simplicity) on $U_{n_0}$ such that for any $x\in U_{n_0}$, $\mu\left(\left\{x\right\}\right) \gt 1/n_0$. Such a measure cannot be $\sigma$-finite. Indeed, a set $S \subset U_{n_0}$ has a finite $\mu$-measure if and only if it is finite, and $U_{n_0}$ is not a countable union of finite sets. 
