An Intermediate Value Theorem for Measures Let $\mu$ be a Lebesgue-Stieltjes measure on R with a continuous distribution function and let A ∈ B(R) with µ(A) > 0. Prove that for each b ∈ (0, µ(A)) there exists a Borel set B ⊆ A such that µ(B) = b. 
I want to use the intermediate value theorem on G(x) = µ (A ∩ [−n, x])  for suitable n.
I have been able to prove it for the case where $\mu (A) < \infty$ by saying:
Consider the sets $K = \{x\in A|x\leq k\}$ and M = $\{x \in A |x \geq m\} $ for $m,k \in N $
Since $\mu (A)$ is finite, there exists $k^*, m^* \in N$ such that $\mu(K)<1/n$, and $\mu(M)<1/p$
Define a function $G:R \rightarrow R $ by 
$ G(x) = \mu $ (A $\cap$ [k*,x])
Then G is continuous since $\mu$ is continuous.
Consider $G(k^*) = \mu $ (A $\cap [k^*,k^*]) < 1/n$ since 
$ A \cap [k^*,k^*] \subseteq K$
$G(m^*) = \mu(A \cap [k^*,m^*]) < \mu(A) - 1/n - 1/p $ 
Setting n = 1/b, and p = $1/(\mu(A)-2b)$ gives us 
$G(k^*) <b$ , $G(m^*) >b$ so by intermediate value theorem there is a c $\in (k^*,m^*)$  s.t. $G(c) = b$
So $\mu(A \cap [k^*,c]) = b, A \cap [k^*,c]$ is a borel set and $A \cap [k^*,c] \subseteq A$
But I don't know what to do if $\mu(A) = \infty$
 A: $\newcommand{\eps}{\varepsilon}$
I came upon a similar idea while trying to solve this problem. Check if this is correct, if not, I'll remove this.
Similar to you, I define $f(x) = \mu(A \cap (-\infty, x])$. Clearly $f$ is monotonic.
First, $\lim_{x \to -\infty} f(x) = 0$. To show this, consider the decreasing sets $(-\infty, 0] \supset (-\infty, -1] \cdots $, which decrease to $\varnothing$. So $f(-n) \to 0$ as $n \to -\infty$.
Similarly, $\lim_{x \to \infty} f(x) = \mu(A)$ for both finite and infinite $\mu(A)$.
Now, lets prove that $f$ is continuous. For this, for fixed $a$ and $\eps > 0$, lets find a $\delta$ such that $|f(x) - f(a)| < \eps$ for all $x \in [a, a + \delta]$. 
To do this, consider the sequence $\delta_n = 1/n$. Then $f(a + \delta_n) - f(a) = \mu(A \cap (a, a + 1/n])$. But the sets $(a, a + 1/n] \searrow \varnothing$. So $f(a + \delta_n) \searrow f(a)$. So there is a $n$ such that $f(a + \delta_n) < f(a) + \eps$. Pick $\delta = \delta_n$.
But note we also have to find such a $\delta$ on the left hand side, i.e., for fixed $a$ and $\eps > 0$, we need to find a $\delta$ such that $|f(x) - f(a)| < \eps$ for all $x \in [a - \delta, a]$. 
Again set $\delta_n = 1/n$. Then $f(a) - f(a - \delta_n) = \mu(A \cap (a - 1/n, a])$. But $(a - 1/n, a] \searrow \{a\}$ (Not $\varnothing$ !). So $f(a - \delta_n) \nearrow f(a)$, as $\lim_{n \to \infty} \mu(A \cap (a - 1/n, a]) = \mu(A \cap \{a\}) = 0$ (it is here that we use the special property of the Lebesgue-Stieltjes measure that $\mu(\textrm{singleton}) = 0$, i.e., $\mu$ is atomless. As you may have observed the measure $\mu$ cannot be something like the Dirac measure).
Now we pick $\delta$ as before, and pick the minimum of the two deltas as our final $\delta$. This completes the proof of the continuity of $f$.
Now we just apply intermediate value theorem to $f$. Let $y$ be such that $f(y) = m$ for $0 < m < \infty$ (we can't pick 0, because it then "$y = -\infty$" may be the only solution). Fixing $0$ is trivial though. Pick $A \cap \mathbb{Q} \subset A$.
Applying IVT to an unbounded domain may be objectionable. We can fix that. Pick $0 < m < \mu(A)$. Then there is an $n$ such that $f(-n) < m < f(n)$ (using the limits for $f$ to $\infty$ and $-\infty$). Apply IVT to $f$ on $[-n, n]$. 
