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?

Solution 1:

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.