# Preservation of convergence in measure by absolutely continuous measures

Mathematics Asked on January 3, 2022

In a paper on Risk theory that I am reading, it is stated that unlike convergence in $$L_p$$, $$1leq p, 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 $$null mu$$. 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.

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)

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 $$mu(|X_n-X|>alpha) Then by the Lemma $$nu(|X_n-X|>alpha) This shows that indeed, $$X_nxrightarrow{nrightarrowinfty}X$$ in $$nu$$-measure.

Short proof of Lemma:

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

Answered by Oliver Diaz on January 3, 2022

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

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

Answered by ir7 on January 3, 2022