Let $\mathcal{M}$ be the set of $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb{R}$. Let $f:\mathbb{R} \rightarrow (0,+\infty)$ be measurable and $g \in L^1(\mathbb{R})$. For $E \in \mathcal{M}$, define \begin{equation*} \mu(E)= \int_E f(x)dx \text{ and } \lambda(E)= \int_E g(x)dx \end{equation*} whenever $E \in \mathcal{M}$. Show $\lambda << \mu$. Find $d\lambda/d\mu$.
if we write f= $\frac{d\mu}{dx}$ and g= $\frac{d\lambda}{dx}$ then $\frac{g}{f} = \dfrac{\frac{d\lambda}{dx}}{\frac{d\mu}{dx}} = \frac{d\lambda}{d\mu}$. So the Radon-Nikodym derivative of $\lambda$ with respect to $\mu$ is $\frac{d\lambda}{d\mu}=\dfrac{g(x)}{f(x)}$.
I need some help to show that $\lambda << \mu$. Given $E \in \mathcal{M}$ such that $\mu(E)=0$, we want to show that $\lambda(E)=0$
If $\mu(E) = 0$, then $f > 0$ implies that $m(E) = 0$, from which it follows that $\lambda(E) = 0$. If you're not comfortable with the differential pushing, you can find the Radon-Nikodym derivative by finding a function $h$ such that $$\lambda(E) = \int_E h \;\mathrm{d}\mu.$$ By uniqueness, such a function is necessarily the Radon-Nikodym derivative. Finding such a function is straightforward: $$\lambda(E) = \int_E g\; \mathrm{d}x = \int_E \frac{g}{f} \cdot f \;\mathrm{d}x = \int_E \frac{g}{f} \;\mathrm{d}\mu.$$ The change of variables in the last step may be justified by approximating with simple functions.
• An ad-hoc term for how OP found the Radon-Nikodym derivative, which involved some shady "cancelling" of d$x$'s that may not be allowed at an introductory level. – user217285 May 16 '17 at 7:04