Construction of Set with required measure I have a Probability measure $P$ defined on $S = [0,1]$ satisfying $P({\{x}\}) = 0$ for any $x$ in $S$.
How can I construct a subset $A$ of $S$ such that $P(A)$ is exactly $\frac{1}{2}$?
Instead of $\frac{1}{2}$, can this construction be generalized to any $0 < k < 1$?

The function $\phi(t) = \mathbb P\big([0,t)\big)$ is continuous.  Therefore it achieves all values between $\phi(0) = 0$ and $\phi(1) = 1$.
As Ramiro remarked, we have to assume $[0,t)$ is $\mathbb P$-measurable for all $t$ to do this.

added: $\phi(t)$ is continuous.
Use countable additivity: since $\phi(t)$ is nondecresing, $$\lim_{h \to 0+}\phi(t+h)= \lim_{h \to 0+}P\big([0,t+h)\big) = \lim_{n \to \infty}P\left([0,t+\textstyle\frac{1}{n})\right) \\
= P\left(\bigcap_{n}[0,t+\textstyle\frac{1}{n})\right) = P\big([0,t]\big)=P\big([0,t)\big)+P\big(\{t\}\big) = \phi(t)$$ and similarly for $\lim_{h \to 0+}\phi(t-h) = \phi(t)$.
