Am I correct with usage of this generalised Dominated Convergence lemma: a sequence $(f_n)$ in $L^1$ on a bounded domain is strongly convergent if and only if $(f_n)$ is uniformly integrable and $(f_n)$ converges in measure, or pointwise almost everywhere?
By Dunford Pettis I know that $(f_n)$ is uniformly integrable if and only if $(f_n)$ is weakly convergent in $L^1$.
My question is, is it possible to deduce $L^p$ convergence of a bounded sequence $(f_n)\subset L^p$, by showing that, for all $\phi\in L^\infty$: $$1. \int |f_n |^p(x) \phi (x) dx\to \int |f |^p(x) \phi (x) dx$$ and $$2. \,\, f_n\to f \text{ in measure} $$ which would imply the uniform integrability of $|f_n-f|^p$, and hence the desired convergence? I would attempt to use the Vitali lemma, so that on a bounded domain $\Omega$, $(|f_n|^p)$ uniformly bounded in $L^1(\Omega)$, existence (and finiteness) of the pointwise a.e limit and uniform integrability of $|f_n-f|^p$ allow passage to the limit $$\lim_n \int_\Omega |f_n-f|^p dx = \int_\Omega\lim_n |f_n-f|^p dx =0.$$
Furthermore, how limited would an approximation argument for 1., with $\phi\in C_c^\infty$? Is there a counterexample with a uniformly integrable sequence, pointwise converging, that does not converge in $L^p$? I ask this second question because of the equivalence of distributional convergence to weak convergence in $L^p$ if the sequence (and limit) are already bounded in $L^p$.
Note: I am using the following definition of UI: $\forall\varepsilon>0, \exists g$ an integrable function such that $$\sup_{n}\int_\Omega(|f_n|-g)^+\mathrm dx\lt\varepsilon.$$
I just noted that the statement you want to prove is actually trivial, because if
$$\int |f_n - f|^p \cdot \phi \, dx \rightarrow 0$$
for all $\phi \in L^\infty$, you can use this with $\phi \equiv 1$ which is then nothing but convergence in $L^p$.
EDIT: If you instead require $\phi \in C_c^\infty(\Omega)$, the statement becomes false in general. As a counterexample consider $p=2$, $\Omega = (0,1)$ and $f_n = \sqrt{n} \cdot \chi_{(0,1/n)}$. Of course, in this case, the sequence is not uniformly integrable.