A question on Lebesgue measure and its absolutely continuous measures Let $\lambda$ be the Lebesgue measure and $\mu$ be a measure absolutely continuous with respect to $\lambda$.
Suppose for Lebesgue measurable sets $X_n$, $n=1,2,...$, 
$\lambda(X_n)=1/n$.
Do we have  $\mu(X_n)\to 0$ as $n\to \infty$?
Thanks.
 A: Consider the measure $\mu(dx) = |x| \, \lambda(dx)$ and $X_n = [n,n+\frac{1}{n}]$. Then $\mu$ is absolutely continuous w.r.t. $\lambda$ but
$$ \mu(X_n) = \int_{X_n} |x| \, dx \geq 1. $$
A: Unless $\mu$ is a finite measure this is, in general, not true. Consider for instance
$$\mu(B) := \int_{B \cap (0,\infty)} x \, \lambda(dx)$$
and
$$X_n := \left[n, n+ \frac{1}{n} \right],$$
then
$$\mu(X_n) = \int_n^{n+1/n} x \, \lambda(dx) \geq n \int_{n}^{n+1/n} \, \lambda(dx) = 1;$$
in particular $\mu(X_n)$ does not converge to $0$.
A: Define $\mu$ on the Lebesgue measurable sets by
$$\mu(E)=\begin{cases}
0 & \text{if}\ \lambda(E)=0,\\
\infty & \text{otherwise}.
\end{cases}$$
The fact that this is a measure follows by using that $\lambda(\emptyset)=0$ and $\lambda(\cup_n E_n)=0$ iff $\lambda(E_n)=0$ for each $n$. Clearly $\mu \ll \lambda$. However $\mu(X_n) = \infty$ for every $n$.
If $\mu$ is assumed to be finite, then it is true. In fact, we have the following theorem (Theorem 3.5 from Folland's Real Analysis).

Theorem. Let $\nu$ be a finite signed measure and $\mu$ a positive measure on $(X,\mathcal{M})$. Then $\nu \ll \mu$ iff for every $\varepsilon>0$ there exists $\delta>0$ such that $|\nu(E)|<\varepsilon$ whenever $\mu(E)<\delta$.

The reverse direction is easy, and the forward direction follows from a standard argument by contradiction using continuity from above.
