Let $p\ge 1$ and suppose that $X_n \to X$ in probability. If there exists $Y\in L^p$ such that $|X_n|\le Y$. Prove that $X_n \to X$ in $L^p$.

Asked Dec 26 '21 at 18:58

Active Dec 26 '21 at 18:58

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Let $p\ge 1$ and suppose that $X_n \to X$ in probability. If there exists $Y\in L^p$ such that $|X_n|\le Y$. Prove that $X_n \to X$ in $L^p$.

I try to prove it as follows. Fix $\epsilon>0$, $$ \int |X_n-X|^p dP=\int_{|X_n-X|\le \epsilon}|X_n-X|^p dP+\int_{|X_n-X|> \epsilon}|X_n-X|^p dP $$ $$ \le \epsilon^p+\int_{|X_n-X|> \epsilon}(2Y)^p dP $$

But I am stuck in there....

asked Dec 26 '21 at 18:58

Hermi

1,480

Dominated convergence along a subsequence? – Dec 26 '21 at 19:03
Can you explain how to use DCT? Thanks! – Hermi Dec 26 '21 at 19:03
https://math.stackexchange.com/questions/3374830/dominated-convergence-theorem-with-almost-surely-replaced-by-convergence-in-p – Dec 26 '21 at 19:08

Let $p\ge 1$ and suppose that $X_n \to X$ in probability. If there exists $Y\in L^p$ such that $|X_n|\le Y$. Prove that $X_n \to X$ in $L^p$.

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