Kalle's super-measure Is there a measure in which the measure of $[0,a)$ is less then $[0,a]$ for all real numbers $a$?  And which is finite on intervals? And we allow the measure to take values of non-real numbers.
 A: Suppose $\mu$ were such a measure.  Then the singletons $\{ a \}$ for $a > 0$ must have positive measure (as $\mu ( [ 0 , a ] ) = \mu ( [ 0 , a ) ) + \mu ( \{ a \} )$).  Thus there is an $n$ such that $\mu ( \{ a \} ) \geq \frac 1n$ for uncountably many $a \in ( 0 , 1 )$ and therefore $\mu ( [ 0 , 1 ) ) = + \infty$.  (Similar arguments can be made for every nondegenerate interval $[0,b)$ ($b > 0$).)

Added: Some looking around regarding the added question of a (possibly) not-real-valued measure with these properties has lead me to the following partial result.
Theorem (Bernstein-Wattenberg, 1969):  There is a finitely-additive hyperreal-valued "measure" $\lambda^*$ on all subsets of $[0,1]$ (the real interval) such that the following hold:


*

*For any Lebesgue measurable $A \subseteq [0,1]$ the standard part of $\lambda^* ( A )$ is $\lambda (A)$ (where $\lambda$ denotes the usual Lebesgue measure);

*Every nonempty set has nonzero $\lambda^*$-measure.


Clearly for such a $\lambda^*$ we have $\lambda^* ( [ 0 , a ) ) < \lambda^* ( [ 0 , a ] )$ for all $0 < a \leq 1$.
A: Suppose there is a measure $\mu$ on $\mathbb{R}$ equipped with some $\sigma$-algebra $\mathcal{B}$ that contains all intervals of the form $[0,a]$ and $[0,a)$ for every $a>0$ such that
$$
\mu([0,a])<\infty,\quad a>0,
$$
and
$$
\mu([0,a))<\mu([0,a]),\quad a>0.
$$
Then 
$$
\mu(\{a\})=\mu([0,a]\setminus [0,a))=\mu([0,a])-\mu([0,a))>0,\quad a>0.
$$
In particular $\mu([0,a])=\infty$ which is contradiction.
A: On $\mathcal{P}(\mathbb{R})$, define
$$\sum_i S_i = \bigcup_i S_i$$
$\mathcal{P}(\mathbb{R})$ is partially ordered by $\subseteq$. In particular, for $a \geq 0$, $[0,a)$ is strictly less than $[0,a]$ in this partial ordering.
We can then define a "measure" by
$$ \mu(S) = S $$
This measure is countably additive (even arbitrarily additive!) because
$$ \mu\left( \bigcup_i  S_i \right) =\bigcup_i  S_i = \sum_i  S_i $$
and this identity even holds when the $S_i$ are not disjoint!
It may be more interesting to consider $\mu(S) = S \cap \mathbb{Z}$.
Of course, it's not clear that this is useful at all....
