Continuous complex measures

Question

I was wondering if following statements are true:

If $\mu$ is the Lebesgue measure and $\mu(A)=\alpha_0$, then it's not difficult to verify that for any $\alpha<\alpha_0$, there exist $B\subset A$ with $\mu(B)=\alpha$.

Then I was thinking if the same is true for any complex Borel measure $\lambda$ absolutely continuous with respect to Lebesgue measure. That is, if $\lambda$ is a complex Borel measure, $\lambda\ll\mu$ and $\lambda(A)=\alpha_0\in\mathbb{R}^+$, is it true that for any $0<\alpha<\alpha_0 $, there exists $B\subset A$ such that $\lambda(B)=\alpha$? If not, does it hold if I add the assumption that $\lambda$ is real valued?

Also I was wondering about generalization of above statement. If $\lambda_1$ and $\lambda_2$ are complex Borel measures and absolutely continuous with respect to Lebesgue measure and $\lambda_1(A)=\lambda_2(A)=\alpha_0\in\mathbb{R}^+$, is it true that for any $0<\alpha<\alpha_0 $, there exist $B\subset A$ such that $\lambda_1(B)=\lambda_2(B)=\alpha$? If not, does it hold if I add the assumption that $\lambda_1$ and $\lambda_2$ are real valued?

It would be great if someone could help me with this, or at least let me know about a reference to read about this.

Answer 1

An important property of Lebesgue measure is that it is nonatomic. A measure $\mu$ on a measurable space $(\Omega,\Sigma)$ is nonatomic if for every $B\in\Sigma$ such that $\mu(B)>0$, there is $A\in\Sigma$ such that $A\subseteq B$ and $0<\mu(A)<\mu(B)$.

A nontrivial $\sigma$-finite Borel measure $\mu$ on $\mathbb{R}$ (or any second countable Hausdorff space for that matter) is nonatomic if and only if $\mu\big(\{x\}\big)=0$ for all $x\in\mathbb{R}$, see here for an sketch of the argument.

So any $\sigma$-finite measure continuous with respect to Lebesgue measure is nonatomic.

If $(\Omega,\Sigma)$ is a measurable space, a finite dimensional vector measure $\nu$ on this measurable space is a $\sigma$-additive function $\nu:\Sigma\to\mathbb{R}^n$ for some $n$. We say that $\nu$ is nonatomic if there is a $\sigma$-finite measure $\mu$ on $(\Omega,\Sigma)$ such that $\lim_{\mu(A_n)\to 0}\nu(A_n)=0$. Such a $\mu$ is a control measure.

Now the Liapounoff convexity theorem says that a finite dimensional vector measure with a finite and nonatomic control measure has compact and convex range. In particular, the range of a finite measure is a closed interval.

Now if $\mu_1$ and $\mu_2$ are $\sigma$-finite Borel measures on $\mathbb{R}$ continuous with respect to Lebesgue measure then both are nonatomic. Suppose $A$ is a Borel set such that $\mu_1(A)=\mu_2(A)=\alpha_0<\infty$. Now consider the measurable space $(A,\mathcal{B}_A)$ where $\mathcal{B}_A$ are the Borel subsets of $A$. Then the restriction of $\mu_1+\mu_2$ to $\mathcal{B}_A$ is a finite nonatomic measure. Define a vector measure $\nu:\mathcal{B}_A\to\mathbb{R}^2$ by letting $\nu(F)=\big(\mu_1(F),\mu_2(F)\big)$. Then $\nu$ is a nonatomic finite dimensional vector measure with control measure $\mu_1+\mu_2$. If $ 0<\alpha<\alpha_0$ then we have $$\alpha=\frac{\alpha_0-\alpha}{\alpha_0}0+\frac{\alpha}{\alpha_0}\alpha_0,$$ $$(\alpha,\alpha)=\frac{\alpha_0-\alpha}{\alpha_0}\nu(\emptyset)+\frac{\alpha}{\alpha_0}\nu(A).$$ Since $\nu$ has convex range, there is $B\in\mathcal{B}_A$ such that $\nu(B)=(\alpha,\alpha)$ and hence $\mu_1(B)=\mu_2(B)=\alpha$.

Being blissfully ignorant of complex measures, I leave that case to others.