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 $0 < k < 1$?
 A: If $P$ has no atoms, then the image of $P$ will be $[0,1]$.
See here and also here
The condition $P(\{x\})=0$ for any $x$ in $S$ is not enough to ensure that $P$ has no atoms.
For a counter-example, consider the probability space $([0,1],\mathcal{M}, P)$ where
$$\mathcal{M} =\{E\subseteq [0,1]: E \text{ or } E^c \text{ is countable} \}$$
and $P(E)=0$ if $E$ is countable and  $P(E)=1$ if $E^c$ is countable.
It is easy to check that $\mathcal{M}$ is a $\sigma$-algebra and that $P$ is a probability measure. In this case, the image of $P$ is just $\{0, 1\}$.
Remark 1: If you assume that

*

*$P(\{x\})=0$ for any $x$ in $S$ and

*For every $x \in [0,1]$, $[0,x) \in \mathcal{M}$
Then
$\mathcal{M}$ contains the Borel $\sigma$-algebra of $[0,1]$ and $P$ has no atoms. So the image of $P$ is $[0,1]$.
Remark 2: As remarked by GEdgar, in the special case presented in Remark 1, a nice way to see that the image of $P$ is $[0,1]$ is to note that the function $x \to P([0,x))$  is continuous , it has value $0$ at $0$ and value $1$ at $1$.
A: 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 $\lim_{h \to 0+}\phi(t-h) = \phi(t)$.
