I need an analysis about the convergence of a sequence of measurable functions. I have written up a proof but I am not sure if it works. Would be grateful if someone could give it a once-over.
Part 1) Assuming that for a given sequence of measurable functions on some measure space $(S, \mathcal{S}, \mu)$ such that $v_n \rightarrow v$ in the measure $\mu$, and for some $r > 0$ we have that: \begin{equation}\int|v_n|^rd\mu\rightarrow\int|v|^rd\mu\end{equation} (these integrals have been assumed to exist), we wish to prove that: \begin{equation} \int|v_n - v|^r d\mu\rightarrow 0 \end{equation} We know that the convergence in measure means that there's a subsequence such that $v_{n_l}\rightarrow v$ $\mu$-a.e.. This implies that $|v_{n_l} - v|^p \rightarrow 0$, for all $p>0$. The idea then is to dominate the sequence $|v_{n_l} - v|^p$ by some $u_{n_l}$, such that $u_{n_l} \rightarrow u$, and $\int u_{n_l} \rightarrow \int u<\infty$ and get the conclusion by the \textbf{generalised Dominated convergence Theorem}. We begin with noticing that for $r\in(0,\infty)$: \begin{equation} |v_{n_l} - v| \le |v_{n_l}| + |v| \implies |v_{n_l} - v|^r \le (|v_{n_l}|+|v|)^r \end{equation} \textbf{Case 1:} We first assume that $r\in[1,\infty)$. Then from the fact that for non-negative $a$ and $b$: $(a+b)^r \le 2^ra^r + 2^rb^r$, and the bound just before:
\begin{equation}|v_{n_l} - v|^r\le 2^r(|v_{n_l}|^r+|v|^r)\end{equation} Now we set $u_{n_l}\overset{\text{def}}{=} 2^r(|v_{n_l}|^r+|v|^r)$. It is obvious that $u_{n_l}\rightarrow 2^{r+1}|v|^r$ $\mu$-a.e., so we define $u\overset{\text{def}}{=} 2^{r+1}|v|^r$. Finally note that: $$\int u_{n_l} = 2^r\left(\int |v_{n_l}|^r + \int |v|^r\right)\rightarrow 2^{r+1}\int|v|^r < \infty$$ \textbf{Case 2:} Assume now that $r\in (0,1)$. Now with the inequality: $(a + b)^r\le a^r + b^r$, and again (5.1), we have: $$|v_{n_l}-v|^r \le |v_{n_l}|^r+|v|^r$$ Finally by setting $u_{n_l}$, as the right hand side of the above and $u\overset{\text{def}}{=} 2|v|^r$, we are able to get the exact same conclusion. We have shown that the hypothesis for the Generalised dominated convergence theorem are met in either case.
Part 2) Given two sequences of measurable functions $v_n$ and $u_n$ on a ameasure space $(S,\mathcal{G},\mu)$ such that $v_n\rightarrow v$ and $u_n \rightarrow u$, and for some conjugate pair of exponents $q, q'\in(1,\infty)$ for which it is true that: $$\int|v_n|^q\;d\mu \rightarrow \int|v|^q\;d\mu \mbox{ and }\int|u_n|^{q'}d\mu\rightarrow \int|u|^{q'}d\mu$$ We want to show that: $$\int |v_nu_n - vu|\;d\mu\rightarrow 0$$ To this end we consider the following bound which comes from the triangle inequality and H"{o}lder's Inequality: \begin{equation*} \begin{split} \int |v_nu_n - vu|\;d\mu &= \int |v_nu_n - v_nu + v_nu - vu|\;d\mu \\ &\le \int |v_nu_n - v_nu|\;d\mu + \int |v_nu - vu|\;d\mu \\ &= \int |v_n||u_n - u|\;d\mu + \int |u||v_n - v|\;d\mu \\ &= \left(\int |v_n|^q\;d\mu\right)^{\frac{1}{q}}\left(\int|u_n - u|^{q'}\;d\mu\right)^{\frac{1}{q'}} + \left(\int |u|^{q'}\;d\mu\right)^{\frac{1}{q'}}\left(\int|v_n - v|^{q}\;d\mu\right)^{\frac{1}{q}} \end{split} \end{equation*} Then the required result follows from taking the limit.