Lebesgue integral and change of measure I have just read the following statement in my textbook:
If $\mu << \nu$, ($\mu$ is absolutely continuous wrt $\nu$), then: 
$$\int g d\mu=\int g \frac{d\mu}{d\nu}d\nu.$$
The only thing I understand in this statement is that because $\mu<<\nu$, if $\nu(A)=0$ for some set A, then $\mu(A)=0$. Can someone explain what it means to change measures and why this notation makes sense?
 A: If $\mu$ and $\nu$ are $\sigma$-finite, then the Radon-Nikodym theoren states that absolute continuity is equivalent to the existence of a density of $\mu$ with respect to $\nu$, i.e. a function $f$ such that
$$
\mu(A)=\int_A f\textrm{d}\nu,
$$
for any measurable set $A$.
This $\nu$-a.e. unique function $f$ is then denoted $\frac{d \mu}{d\nu}$ and it follows more or less immediately (by applying the standard machine of considering indicators, then simple functions, then applying monotone convergence and then linearity), that
$$
\int g\textrm{d}\mu=\int gf\textrm{d}\nu
$$
for any $g\in L^1(\mu)$.
Note that the existence of a density is false without the $\sigma$-finite assumption. Take for instance $\nu$ to be the counting measure on $\mathbb{R}$, then $\mu <<\nu$ for any measure $\mu$, since $\nu(A)=0$ implies $A=\emptyset$. However, if $\mu$ is the Lesbegue measure, then there exists no $f$ which can play the role of density, since it easily follows that for any $f\geq 0$ and any Borel set $A$
$$
\int_A f\textrm{d}\nu=\sum_{x\in A} f(x),
$$
which is only non-zero if there exists $x_0$ in $A$ such that $f(x_0)\neq 0$. However, then,
$$
\mu(\{x_0\})=\int_{\{x_0\}} f\textrm{d}\nu= f(x_0)\neq 0,
$$
and the Lesbegue measure doesn't have point masses.
