Preservation of convergence in measure by absolutely continuous measures

This statement can be obtained as a consequence of the following Lemma:

Lemma: If $nullmu$ and $nu$ is finite (as is in your case) then for any $varepsilon>0$, there is $delta>0$ such that for any $Ainmathcal{F}$, $$ nu(A)<deltaquadtext{implies}quadmu(A)<varepsilon $$

A leave a short proof of this at the end of this answer.

To apply the Lemma to your situation, fix $alpha>0$ and $varepsilon>0$. Let $delta>0$ be as in the Lemma. Since $X_nxrightarrow{nrightarrowinfty}X$ in $mu$ measure, there is $n_0inmathbb{N}$ such that $$ mu(|X_n-X|>alpha)<deltaqquad ngeq n_0 $$ Then by the Lemma $$ nu(|X_n-X|>alpha)<varepsilonqquad ngeq n_0 $$ This shows that indeed, $X_nxrightarrow{nrightarrowinfty}X$ in $nu$-measure.

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Short proof of Lemma:

$Longrightarrow$: Suppose that for any $varepsilon>0$, there is $delta>0$ such that $|nu(A)|<varepsilon$ whenever $Ainmathcal{F}$ and $mu(A)<delta$.
If $mu(E)=0$ then $nu(E)<varepsilon$ for all $varepsilon>0$; consequently $nu(E)=0$. This means that $nullmu$.

$Longleftarrow$: In the other direction, suppose that there exist $varepsilon>0$ for which there is a sequence ${A_n}subsetmathscr{F}$ with $mu(A_n)<2^{-n}$ but $nu(A_n)geq varepsilon$. Define $A=bigcap_nbigcup_{mgeq n}A_m$. Clearly $mu(A)=0$. however, $$ infty>nu(Omega)geq nu(A)=lim_nnu(bigcup_{mgeq n}A_m)geqliminf_nnu(A_n)geqvarepsilon. $$ Which means that $nu$ is not absolutely continuous with respect to $mu$.

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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

$$ nu({omega in Omega: X_{n_{k_h}}(omega) mathrm{; does; not ; converge ; to ;} X(omega) }) $$