Tricky Lebesgue Dominated Convergence Question Let $f, f_k\in L^p$, $1<p<\infty$, such that $\|f_k\|_{L^p(\mathbb{R}^n)}\leq M$ for all $k$, where $0<M<\infty$.
If $f_k\to f$ a.e., prove that $$\large\lim_{k\to\infty}\int|f_k|^{p-2}f_kg\,dx=\int|f|^{p-2}fg\,dx$$
for all $g\in L^p(\mathbb{R}^n)$.

My attempt:
My first thought would be to try Lebesgue's Dominated Convergence Theorem.
We have $||f_k|^{p-2}f_kg|=|f_k|^{p-1}|g|$ which turns out to be actually integrable by Holder's inequality:
\begin{align*}
\int|f_k|^{p-1}|g|&\leq\|f_k^{p-1}\|_q\|g\|_p\\
&=(\int|f_k|^p)^{1/q}\|g\|_p<\infty
\end{align*}
where $q=p/(p-1)$.
However, I can't find a dominating function here.
Thanks for any help.

I also know this theorem called "Converse of Holder's inequality", but am not sure if it is useful here:
$\|f\|_p=\sup\int fg$, where the supremum is taken over all real-valued $g$ such that $\|g\|_{p'}\leq 1$ and $\int fg$ exists.
 A: Upon further thinking, I don't see how one could use the dominated convergence theorem (standard or generalised) here. So let's take a different route.
Lemma: Let $1 \leqslant q \leqslant \infty$, and $p$ the conjugate exponent to $q$ (that is, $\frac{1}{q} + \frac{1}{p} = 1$). Let $(h_n)_{n \in \mathbb{N}}$ a sequence in $L^q(\mu)$, where $\mu$ is a $\sigma$-finite measure, such that


*

*$h_n \to w$ with respect to $\sigma(L^q(\mu),L^p(\mu))$ (that is weak convergence for $1 \leqslant q < \infty$ and weak$^\ast$ convergence for $1 < q \leqslant \infty$), and

*$h_n \to h$ pointwise almost everywhere.


Then $h = w$ almost everywhere.
Proof: We first note that $h \in L^q(\mu)$ and $\lVert h\rVert_{L^q(\mu)} \leqslant \sup \:\bigl\{ \lVert h_n\rVert_{L^q(\mu)} : n \in \mathbb{N}\bigr\}$. Since $L^p(\mu)$ is a Banach space, the Banach-Steinhaus theorem yields $S := \sup \:\bigl\{ \lVert h_n\rVert_{L^q(\mu)} : n \in \mathbb{N}\bigr\} < \infty$. For $q < \infty$, Fatou's lemma and the a.e. convergence yield
$$\int \lvert h\rvert^q\,d\mu = \int\liminf_{n\to\infty} \lvert h_n\rvert^q\,d\mu \leqslant \liminf_{n\to\infty} \int \lvert h_n\rvert^q\,d\mu \leqslant S^q,$$
and for $q = \infty$, there is a $\mu$-null set $N$ such that $h_n(x) \to h(x)$ for $x\notin N$, and $\lvert h_n(x)\rvert \leqslant S$ for all $n\in \mathbb{N}$ and all $x\notin N$. Then we also have $\lvert h(x)\rvert = \lim_{n\to\infty} \lvert h_n(x)\rvert \leqslant S$ for $x\notin N$.
Assume $\mu(\{ x : h(x) \neq w(x)\}) > 0$. By multiplying with a suitable power of $i$, we can assume that $A := \{ x : \operatorname{Re} h(x) > \operatorname{Re} w(x)\}$ is not a null set. Since $\mu$ is $\sigma$-finite, there is a measurable $B \subset A$ with $0 < \mu(B) < +\infty$. By Egorov's theorem, there is a measurable $C\subset B$ with $\mu(C) > 0$ such that $h_n$ converges uniformly to $h$ on $C$. Then $\chi_C \in L^p(\mu)$, and
$$\operatorname{Re} \int h(x)\chi_C(x)\,d\mu > \operatorname{Re} \int w(x)\chi_C(x)\,d\mu\tag{$\ast$}$$
by the choice of $C$. But we have
$$\lim_{n\to\infty} \int h_n(x)\chi_C(x)\,d\mu = \int w(x)\chi_C(x)\,d\mu$$
by the assumed $\sigma(L^q(\mu), L^p(\mu))$ convergence, and
$$\lim_{n\to\infty} \int h_n(x)\chi_C(x)\,d\mu = \int h(x)\chi_C(x)\,d\mu$$
by the uniform convergence on $C$ and the finiteness of $\mu(C)$, and this contradicts $(\ast)$. So the assumption $\mu(\{ x : h(x) \neq w(x)\}) > 0$ must have been wrong.
Corollary: Let $\mu$ a $\sigma$-finite measure, and $1 < q < \infty$. If $(h_n)_{n\in \mathbb{N}}$ is a bounded sequence in $L^q(\mu)$ that converges pointwise almost everywhere to $h$, then $h \in L^q(\mu)$, and $(h_n)$ converges weakly to $h$.
Proof: Since $L^q(\mu)$ is reflexive, every bounded sequence has a weakly convergent subsequence. By the lemma, every subsequence $(h_{n_k})_{k\in \mathbb{N}}$ of $(h_n)$ has a further subsequence $\bigl(h_{n_{k_m}}\bigr)_{m\in \mathbb{N}}$ that converges weakly to $h$. But that means the full sequence converges weakly to $h$.
Applying the corollary to $h_k = \lvert f_k\rvert^{p-2}f_k$ and $h = \lvert f\rvert^{p-2}f$ yields the conclusion.
