Let $f_{n} \subseteq L^{p}(X, \mu)$, $1 < p < \infty$, which converge almost everywhere to a function $f$ in $L^{p}(X, \mu)$ and suppose that there is a constant $M$ such that $||f_{n}||_{\infty} \leq M$ for all $n$. Then for each function $g \in L^{q}(X, \mu)$ (where $q^{-1} + p^{-1} = 1$) we have $\lim_{n\to \infty} \int f_{n}g = \int fg$.
I am able to prove this via two different routes, although they are each short they rely on a number of previous results so that the whole business does not look like it is self-contained at all. I guess it depends on what the examiners would want reproduced or else some problems are just about listing previous results in a certain order, am I misguided?
Proof (1):
Step 1: By applying $f = \operatorname{sgn}(g)\left(\frac{|g|}{||g||_{q}} \right)^{q-1}$, and then Hölder's inequality to the operator $\mathcal{F}_{g}(f) := \int fg$, it follows that it is a bounded linear functional.
Step 2: A bounded linear functional is uniformly continuous.
Step 3: A continuous function preserves limits.
Step 4: If we can show that $\lim_{n \to \infty}||f_{n} - f|| = 0$, then we will have the desired result by invoking Step 3.
Step 5: We will use the fact: if every subsequence $y_{n_{k}}$ has a subsequence $y_{n_{k_{l}}} \to y$, then $y_{n} \to y$.
Step 6: Pick a subsequence $f_{n_{k}}$. Since it is bounded by the hypothesis and $L^{p}(X, \mu)$ is a reflexive space it must have a subsequence $f_{n_{k_{l}}} \to f$ and so by Step 5 $f_{n} \to f$ in norm and then the result follows as it was promised.
Proof (2): A variant of this uses the hypothesis to deduce that the function $h(x) = (f - f_{n})^{p}$ meets the hypotheses of the Dominated Convergence Theorem and this together with the Hölder's inequality give the desired result.
If $\mu(X)=\infty$, the result is not true in general. Let $X=[1,\infty)$ with $\mu$ Lebesgue measure and let $f_n$ be the characteristic function of the interval $[n,2\,n]$. Then $f_n\in L^p$, $\|f_n\|_\infty=1$ for all $n$ and $f_n(x)$ converges point wise to $f(x)=0$ for all $x\in X$. Let $g(x)=1/x$. Then $g\in L^q$ for $q>1$. We have $$\int_X f\,g\,d\mu=0\quad\text{but}\int_X f_n\,g\,d\mu=\log2\quad\forall n. $$