When are two measures absolutely continuous?

Question

Let $P, Q$ be two positive $\sigma-$finite measures which are each of them absolutely continues in respect to a third measure $L, P \ll L, Q \ll L.$ Based on this, can one state anything about the absolute continuity of $P$ in respect to $Q$ and vice versa? In particular, let $dP=f_1dL,dQ=f_2dL.$ Then, $\frac{dP}{dQ}=\frac{f_1}{f_2}$ and $\frac{dQ}{dP}=\frac{f_2}{f_1}.$ Can one conclude from here anything about the mutual absolute continuity of $P$ and $Q$? Can one apply here the Radon-Nikodym theorem in stating that the Radon-Nikodym derivative of $P$ in respect to $Q$ is $dP/dQ=f_1/f_2$ and similarely for the Radon-Nikodym derivative of $Q$ in respect to $P$?

Answer 1

For definiteness, I'll look at only one direction. So we have $ P \ll L $ with $ \mathrm d P = f _ 1 \, \mathrm d L $, and $ Q \ll L $ with $ \mathrm d Q = f _ 2 \, \mathrm d L $; and we want to know whether $ P \ll Q $ with $ \mathrm d P = \frac { f _ 1 } { f _ 2 } \, \mathrm d Q $.

Claim: If $ f _ 2 \ne 0 $ almost everywhere (with respect to $ L $), then $ P \ll Q $.

Proof: Let $ E $ be a measurable set, and suppose that $ Q ( E ) = 0 $; then $ \int _ E f _ 2 \, \mathrm d L = 0 $. Let $ E ' $ be $ E \cap \{ f _ 2 \ne 0 \} $, so that $ \int _ E f _ 2 \, \mathrm d L = \int _ { E ' } f _ 2 \, \mathrm d L $, so this is $ 0 $ too. Since $ Q $ is a positive measure (although all of this can be generalized to signed measures by looking at $ Q ^ + $ and $ Q ^ - $), we know that $ f _ 2 \geq 0 $ almost everywhere; since we've assumed that $ f _ 2 \ne 0 $ almost everywhere, we actually have $ f _ 2 > 0 $ almost everywhere. This tells us two important facts about $ E ' $: the first is that $ f _ 2 > 0 $ on $ E ' $; the second is that $ E $ and $ E ' $ are equivalent with respect to $ L $ (in that their symmetric difference is $ L $-null). By the first fact, $ \int _ { E ' } f _ 2 \, \mathrm d L = 0 $ tells us that $ L ( E ' ) = 0 $; by the second fact, this means that $ L ( E ) = 0 $ too. Since $ P \ll L $, we also have $ P ( E ) = 0 $. Therefore, $ P \ll Q $. ∎

(This really proves that $ L \ll Q $; we get $ P \ll Q $ because $ P \ll L \ll Q $.)

Claim: If $ P \ll Q $, and $ g $ is a function that agrees with $ f _ 1 / f _ 2 $ at all points where $ f _ 2 \ne 0 $, then $ \mathrm d P = g \, \mathrm d Q $.

Proof: Let $ E $ be a measurable set. Then $ \int _ E g \, \mathrm d Q = \int _ E g f _ 2 \, \mathrm d L $. As before, let $ E ' $ be $ E \cap \{ f _ 2 \ne 0 \} $, so that $ \int _ E g f _ 2 \, \mathrm d L = \int _ { E ' } g f _ 2 \, \mathrm d L $. Since $ g = f _ 1 / f _ 2 $ on $ E ' $, we can continue: $ \int _ { E ' } g f _ 2 \, \mathrm d L = \int _ { E ' } f _ 1 \, \mathrm d L = P ( E ' ) $. Now, $ Q ( E \setminus E ' ) = \int _ { E \setminus E ' } f _ 2 \, \mathrm d L = 0 $, since $ f _ 2 = 0 $ on $ E \setminus E ' $; since $ P \ll Q $, we also have $ P ( E \setminus E ' ) = 0 $. So $ P ( E ) = P ( E ' ) + P ( E \setminus E ' ) = P ( E ' ) $. In summary, we have $ P ( E ) = P ( E ' ) = \int _ { E ' } g f _ 2 \, \mathrm d L = \int _ E g f _ 2 \, \mathrm d L = \int _ E g \, \mathrm d Q $ for any measurable $ E $; so $ \mathrm d P = g \, \mathrm d Q $. ∎

Combining these claims, we get that if $ f _ 2 \ne 0 $ almost everywhere, then $ P \ll Q $ with $ \mathrm d P = \frac { f _ 1 } { f _ 2 } \, \mathrm d Q $, where $ f _ 1 / f _ 2 $ may be defined however you like where $ f _ 2 $ is zero. (But we actually get two slightly stronger claims: if $ f _ 2 \ne 0 $ almost everywhere, then $ L \ll Q $, not just $ P $; and regardless of how much $ f _ 2 \ne 0 $, as long as $ P \ll Q $ still holds, then $ \mathrm d P = \frac { f _ 1 } { f _ 2 } \, \mathrm d Q $.)

Incidentally, we don't really need these measures to be σ-finite, as long as the Radon–Nikodym derivatives $ f _ 1 $ and $ f _ 2 $ exist (which is always possible). And as I remarked earlier, we don't need the measures to be positive either, although the proof of the first claim is a little more involved without that assumption.