As the title says I am looking for the measure theory counterparts of slutsky and continuous mapping theorem. So far I only found both theorems in probability theorey related literature (even for literature that focuses mainly on measure theory). So I wonder, a) are there counterparts at all and if yes b) what additional assumption do they require?
Measure Theory Counterpart of Slutsky and Continuous Mapping Theorem
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probability-theory
measure-theory
1 Answers
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It is an interesting question indeed. Perhaps this article gives a counterpart of sorts for the continuous mapping theorem. It says that convergence almost everywhere is preserved under a continuous mapping (Theorem 1), while convergence in measure is preserved under a uniformly continous mapping (Theorem 2). I am not sure how we would go about convergence in distribution as it is a probabilistic term.
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0Thanks for the article. There exists the concepts of convergence of measures in measure theory (see https://en.wikipedia.org/wiki/Convergence_of_measures) which is the counterpart to convergence in distribution. – 2017-01-20