Preservation of convergence in measure by absolutely continuous measures In a paper on Risk theory that I am reading, it is stated that unlike convergence in $L_p$, $1\leq p<\infty$, converges in measure is preserved within a collection of probability measures that are absolutely continuous. That is,
Suppose $\mu$ and $\nu$ are probability measures on a measurable space $(\Omega,\mathcal{F})$ and $\nu\ll \mu$. If  the sequence $X_n$ of random variables converging to $X$ in $\mu$-measure,   then $X_n$  converges to $X$ in $\nu$-measure.
This seems to be an easy enough problem, but I don't have a clear idea of how to start. I would appreciate any hints.
 A: Hint:
For finite measures, use this wiki fact: A sequence $X_n$ converges to $X$ in measure if and only if for any subsequence $X_{n_k}$ there is a sub-subsequence $X_{n_{k_h}}$ that converges to $X$ almost everywhere.
Then compare (for the relevant sub-subsequence):
$$ \mu(\{\omega \in \Omega: X_{n_{k_h}}(\omega) \mathrm{\; does\; not \; converge \; to \;} X(\omega) \}) $$
and
$$ \nu(\{\omega \in \Omega: X_{n_{k_h}}(\omega) \mathrm{\; does\; not \; converge \; to \;} X(\omega) \}) $$
A: This statement can be obtained as a consequence of the following Lemma:

Lemma: If $\nu\ll\mu$ and $\nu$ is finite (as is in your case) then for any $\varepsilon>0$, there is $\delta>0$ such that for any
$A\in\mathcal{F}$, $$
 \nu(A)<\delta\quad\text{implies}\quad\mu(A)<\varepsilon $$

A leave a short proof of this  at the end of this answer.
To apply the Lemma  to  situation described in the problem, let us fix $\alpha>0$ and $\varepsilon>0$. Let $\delta>0$ be as in the Lemma.  Since $X_n\xrightarrow{n\rightarrow\infty}X$ in $\mu$ measure, there is $n_0\in\mathbb{N}$ such that
$$
\mu(|X_n-X|>\alpha)<\delta\qquad n\geq n_0
$$
Then by the Lemma
$$
\nu(|X_n-X|>\alpha)<\varepsilon\qquad n\geq n_0
$$
This shows that indeed, $X_n\xrightarrow{n\rightarrow\infty}X$ in $\nu$-measure.

Short proof of Lemma:
$\Longrightarrow$: Suppose that for any $\varepsilon>0$, there is $\delta>0$
such that $|\nu(A)|<\varepsilon$ whenever  $A\in\mathcal{F}$ and
$\mu(A)<\delta$.
If $\mu(E)=0$ then
$\nu(E)<\varepsilon$ for all $\varepsilon>0$; consequently $\nu(E)=0$. This means that $\nu\ll\mu$.
$\Longleftarrow$: In the other direction, suppose that there exist $\varepsilon>0$ for which there is a sequence
$\{A_n\}\subset\mathscr{F}$ with $\mu(A_n)<2^{-n}$ but $\nu(A_n)\geq \varepsilon$. Define $A=\bigcap_n\bigcup_{m\geq n}A_m$.  Clearly $\mu(A)=0$. however,
$$
\infty>\nu(\Omega)\geq \nu(A)=\lim_n\nu(\bigcup_{m\geq
  n}A_m)\geq\liminf_n\nu(A_n)\geq\varepsilon.
$$
Which means that $\nu$ is not absolutely continuous with respect to $\mu$.

