If $X_n(\omega )\overset{L^2}{\rightarrow}X(\omega )$ and g(x) is continuous and bounded so $g(X_n(\omega ))\overset{a.s.}{\rightarrow}g(X(\omega ))$

Asked Jul 07 '22 at 17:34

Active Jul 07 '22 at 17:34

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Let a sequence of r.v. $X_n(\omega )\overset{L^2}{\rightarrow}X(\omega )$ and $g(x)$ a continuous bounded function.
Is this implies ,and if yes can you please give the prove, that $g(X_n(\omega ))\overset{a.s.}{\rightarrow}g(X(\omega ))$

asked Jul 07 '22 at 17:34

X0-user-0X

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What did you try? Add that in an edit. – Snoop Jul 07 '22 at 18:52
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Does this answer your question? Does convergence in $L^p$ imply convergence almost everywhere? – Math1000 Jul 08 '22 at 23:32
@Math1000 Thanks for your answer but NO i allready know that for exemple the typewriter sequence fct is a counter exemple that converge in L2 but doesn't converge point wise (no where).
I m speaking in a special case when g(x) is bounded. I ve yesterday heard about this theorem https://en.wikipedia.org/wiki/Dominated_convergence_theorem that can help me answer my question. But sadly i ve an exam soon and i can't spend too much time on it. I ll maybe try to writte a full and details answer to my question (based on this theorem) when i ll have more time, and if i succeed to do it of course. – X0-user-0X Jul 09 '22 at 08:52
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Oh, I missed that detail. Close vote retracted. – Math1000 Jul 10 '22 at 20:44
Somebody can help me? – X0-user-0X Jul 12 '22 at 18:12
Anyone to help me? – X0-user-0X Jul 16 '22 at 08:36

If $X_n(\omega )\overset{L^2}{\rightarrow}X(\omega )$ and g(x) is continuous and bounded so $g(X_n(\omega ))\overset{a.s.}{\rightarrow}g(X(\omega ))$

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