# Convergence in measure implies convergence in $L^p$ under the hypothesis of domination

Given a sequence $f_n \in L^p$ and $g \in L^p$, with $|f_n| \leq g$, I am trying to show that $f_n \to f$ in measure implies $f_n \to f$ in $L^p$.

Firstly, I know that if $f_n \to f$ in measure, then there is a subsequence $f_{n_i}$ such that $f_{n_i} \to f$ almost everywhere. Then I can use the dominated convergence theorem to show that $\lVert f_{n_i} - f_p\rVert \to 0$.

Now I am trying to show that $\lVert f_n - f\rVert_p \to 0$. My idea is to assume that $\lVert f_n - f\rVert_p \nrightarrow 0$ and then show that this contradicts the fact that $\lVert f_{n_i} - f\rVert_p \to 0$, but I am not sure of the details. Can anyone help me finish the argument?

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In the title, you should add "with an hypothesis of domination". You are almost done: if $||f_n-f||_p$ doesn't converge to $0$, then we can find $\delta>0$ and a subsequence such that $||f_{n_k}-f||\geq \delta$. This subsequence still converges in measure to $f$, so by your previous argument we get a contradiction. – Davide Giraudo Oct 18 '12 at 19:01
@DavideGiraudo Ok, got it. Thanks! – rt93 Oct 18 '12 at 19:16
You can answer your own question (hence it won't remain unanswered), and you will have your homework done properly. – Davide Giraudo Oct 18 '12 at 19:17

If $\lVert f_n - f\rVert_p \nrightarrow 0$, there exist a subsequence $f_{n_i}$ such that $\| f_{n_i} - f\| _p \ge \varepsilon$ for some $\varepsilon >0$. But $f_{n_i}$ still converges in measure to $f$. So, again There is a subsequence $f_{n_{i_j}}$ of the $f_{n_i}$ such that $\|f_{n_{i_j}} - f\|_p \rightarrow 0$, and this is a contradiction.