Real Analysis, Folland Corollary 3.10 The Lebesgue Radon Nikodym Theorem Background Information:

Proposition 3.9 - Suppose that $\nu$ is a $\sigma$-finite measure and $\lambda$ are $\sigma$-finite measures on $(X,M)$ such that $\nu\ll \mu$ and $\mu\ll \lambda$.
a.) If $g\in L^1(\nu)$, then $g(d\nu/d\mu)\in L^1(\nu)$ and $$\int g d\nu = \int g \frac{d\nu}{d\mu}d\mu$$
b.) We have that $\nu\ll \lambda$, and $$\frac{d\nu}{d\lambda} = \frac{d\nu}{d\mu}\frac{d\mu}{d\lambda}  \quad \lambda-\text{a.e.}$$

Question:

Corollary 3.10 - If $\mu \ll \lambda$ and $\lambda \ll \mu$ then $$\left(\frac{d\lambda}{d\mu}\right)\left(\frac{d\mu}{d\lambda}\right) = 1 \quad\text{a.e.}$$ with respect to either $\lambda$ or $\mu$.

Attempted proof - Suppose $\mu\ll\lambda$ then by proposition 3.9 b.) $$\frac{d\mu}{d\lambda} = \frac{d\mu}{d\nu}\frac{d\nu}{d\lambda} \quad \lambda- \text{a.e.}$$ Similarly if $\lambda \ll \mu$ then by proposition 3.9 b.) $$\frac{d\lambda}{d\mu} = \frac{d\lambda}{d\nu}\frac{d\nu}{d\mu} \quad \mu- \text{a.e.}$$ Then putting these two quantities together and using the result of proposition 3.9 we have that $$\left(\frac{d\mu}{d\lambda}\right)\left(\frac{d\lambda}{d\mu}\right) = 1 \quad\text{a.e.}$$
I am not sure if this is correct, any suggestions is greatly appreciated.
 A: Your work looks correct.  You can simplify your argument by letting $\lambda$ play the role of $\nu$ in the proposition.  This gives us
$$ \left(\frac{d\mu}{d\lambda}\right)\left(\frac{d\lambda}{d\mu}\right) = \frac{d\lambda}{d\lambda}=1\qquad (\lambda-\textrm{a.e.}). $$
And, by exchanging the roles of $\mu$ and $\lambda$, we have
$$ \left(\frac{d\lambda}{d\mu}\right)\left(\frac{d\mu}{d\lambda}\right) = \frac{d\mu}{d\mu}=1 \qquad (\mu-\textrm{a.e.}).$$
A: @Wolfy , you are taking the longer path and actually using 3.10 to prove 3.10 (when combining the equations you have you seem to be implicitly using 3.10 to simply the result to 1). The correct proof is simpler. 

Corollary 3.10 - If $\mu \ll \lambda$ and $\lambda \ll \mu$ then $$\left(\frac{d\lambda}{d\mu}\right)\left(\frac{d\mu}{d\lambda}\right) = 1 \quad\text{a.e.}$$ with respect to either $\lambda$ or $\mu$.

Proof - Suppose $\lambda \ll \mu$ and $\mu\ll\lambda$.  Then, by proposition 3.9, taking $\nu=\lambda$ , we have  
$$\frac{d\lambda}{d\lambda} = \frac{d\lambda}{d\mu}\frac{d\mu}{d\lambda} \quad \lambda-\text{a.e.}$$
On the other hand, by the last part (last sentence) of theorem 3.8, we have 
$$\frac{d\lambda}{d\lambda} = 1\quad \lambda-\text{a.e.}$$
Combining both equations, we have 
$$\left(\frac{d\lambda}{d\mu}\right)\left(\frac{d\mu}{d\lambda}\right) = 1 \quad \lambda-\text{a.e.}$$
Remark: since $\lambda \ll \mu$ and $\mu\ll\lambda$, we have that, for any measurable set $E$, $\lambda(E)=0$ iff $\mu(E)=0$, so the last equation can be simply written as 
$$\left(\frac{d\lambda}{d\mu}\right)\left(\frac{d\mu}{d\lambda}\right) = 1 \quad \text{a.e.}$$
with no risk of confusion. 
