Proving $\int f_n gd \mu \to \int f g d\mu$ for $f \in L^p, g \in L^q$ where $p$ and $q$ are Holder conjugate exponents. I usually have a hard time with proofs about convergence, so I want to make sure if I did it right or not. The problem is:
If $f_n \to f$ in $L^p$, $g \in L^q$, and $\frac{1}{p} + \frac{1}{q} = 1$, then $$\int f_n g d\mu \to \int fg d\mu$$
By Hölder's inequality, for each $n$,
$$\int |f_n g| d\mu \leq ||f_n||_p ||g||_q < \infty$$
Since $f_n \to f$ in $L^p$, that means that $f$ is integrable and, therefore, measurable. Since $g \in L^q$, it is integrable.
Therefore, by the Dominated Convergence theorem
$$\lim_{n\to \infty}\int f_n g d\mu = \int\lim_{n\to \infty}(f_ng )d\mu = \int fgd\mu$$
 A: You implicitly said that $f,g\in L^{p}$ implies that they are integrable, this is false in general. Being integrable means that $f$ (or $g$) belongs to $L^{1}$. We know that $L^{p}$ functions do not necessarily belong to $L^{1}$ unless the finite measure is in issue.
To use the Lebesgue Dominated Convergence Theorem, you need an integrable upper bound, that is, are you able to find an $L^{1}$ function $h$ such that $|f_{n}(x)g(x)|\leq h(x)$? As commented by @ABP there.
In this case, I cannot see any of such an integrable upper bound, so a simpler way to do it is to do it directly:
\begin{align*}
\left|\int f_{n}g-\int fg\right|=\left|\int(f_{n}-f)g\right|\leq\int|f_{n}-f||g|\leq\|f_{n}-f\|_{L^{p}}\|g\|_{L^{q}}\rightarrow 0
\end{align*}
as $n\rightarrow\infty$. Note that $\|g\|_{L^{q}}$ is just a constant.
A: There is a way to demonstrate a slightly weaker result using Lebesgue dominated convergence, although the problem is already solved by @user284331 I'll share this prove, I guess it'll help you to understand how to proceed with this kind of problem (and the argument is useful with some applications). 
As you have $f_{n} \to f$ in $L^{p}$ you can extract a subsequence $(f_{n_{k}})$ such that 
(i) $f_{n_{k}}(x) \to f(x)$ a.e.;
(ii) $|f_{n_{k}}(x)| \leq h(x)$ a.e. with $h \in L^{p}$.
(you can see Brezis' functional analysis book for a prove of that) so, you have that 
(iii) $f_{n_{k}}(x)g(x) \to f(x)g(x)$ a.e.;
(iv) $|f_{n_{k}}(x)g(x)| \leq h(x)|g(x)|$ a.e. and, in this case, $h|g| \in L^{1}$.
So applying $(iii), (iv)$ you're able to conclude that $\int{f_{n_{k}}g} \to \int{fg}$. This isn't what you want, but this is how you'd have to argue if you want to apply LDCT.
