Let $\mu_1$ and $\mu_2$ be finite positive measures on $(X,\Sigma)$. Show that there exist disjoint measurable sets $A\cup B=X$ such that $\mu_1 \bot \mu_2$ on $(A,\Sigma \cap A)$ and $\mu_1 \ll \mu_2 \ll \mu_1 $ on $(B,\Sigma \cap B)$.

Hint: show that $\mu_1 , \mu_2 \ll \mu_1 + \mu_2 $ and apply Radon-Nikodym theorem.

I couldn't solve this question. I'm certain it's been asked before but I don't know how to look for it. any hints or links would be appreciated.


Following the hint, let $f_1=\frac{d\mu_1}{d(\mu_1+\mu_2)}$, $f_2=\frac{d\mu_2}{d(\mu_1+\mu_2)}$.

Then let $Z_1=f_1^{-1}(0)$, $Z_2=f_2^{-1}(0)$, and let $A=Z_1\cup Z_2$, $B=A^C$.

By construction, $A=Z_1\cup Z_2$, and $\mu_1(Z_1)=0=\mu_2(Z_2)$, so $\mu_1\perp \mu_2$ on $A$.

Thus, you just need to show that $\mu_1\ll \mu_2\ll \mu_1$ on $B$. By symmetry, it suffices to show that $\mu_1\ll \mu_2$.

Suppose that $\mu_2(C)=0$ for some $C\subseteq B$. Now let $C_n = f_2^{-1}(1/n,\infty)$, and consider $$0=\mu_2(C) \ge \mu_2(C_n)=\int_{C_n} f_2 d(\mu_1+\mu_2) \ge \frac{1}{n} (\mu_1+\mu_2)(C_n).$$ Thus $(\mu_1+\mu_2)(C_n)=0$ for all $n$, so $\mu_1(C_n)=0$ for all $n$. However, $f_2(x)\ne 0$ for all $x\in C$, since $C\subseteq B$, so we have that $$ C=\bigcup_{n=1}^\infty C_n,$$ since for any $x\in C$, for some large enough $n$, $f_2(x) > \frac{1}{n}$. Thus $\mu_1(C)=0$, as desired.

  • $\begingroup$ Thanks a lot! When you wrote $D_n$ you meant $C_n$ right? also, shouldn't the integral be on $d(\mu_1 + \mu_2) / f_2$? $\endgroup$ – SlyxBrd Feb 11 at 17:00
  • $\begingroup$ @SlyxBrd whoops yeah there are some typos, I'll fix those when I get back to a computer $\endgroup$ – jgon Feb 11 at 17:03
  • $\begingroup$ Thank you, if you can explain why $\mu_2(C_n)=0$ and maybe elaborate on why $C=\cup C_n$ i would be really grateful, I don't understand these parts. $\endgroup$ – SlyxBrd Feb 11 at 17:29
  • $\begingroup$ @SlyxBrd I've edited $\endgroup$ – jgon Feb 11 at 18:05
  • $\begingroup$ Thank you. About the part where we need to show that $\mu_1,\mu_2 << \mu_1 + \mu_2$, can I say that: $0=( \mu_1 + \mu_2)(A)= \mu_1(A) + \mu_2(A)$ And since they are positive we get $\mu_1(A)=0,\mu_2(A)=0$? $\endgroup$ – SlyxBrd Feb 12 at 10:52

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