# Corollary of Radon-Nikodym theorem

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.

## 1 Answer

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.

• Thanks a lot! When you wrote $D_n$ you meant $C_n$ right? also, shouldn't the integral be on $d(\mu_1 + \mu_2) / f_2$? – SlyxBrd Feb 11 at 17:00
• @SlyxBrd whoops yeah there are some typos, I'll fix those when I get back to a computer – jgon Feb 11 at 17:03
• 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. – SlyxBrd Feb 11 at 17:29
• @SlyxBrd I've edited – jgon Feb 11 at 18:05
• 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 $\mu_1(A)=0,\mu_2(A)=0$? – SlyxBrd Feb 12 at 10:52