I'm trying to prove $f_kg_k \to fg$ in measure if $f_k\to f$ in measure, $g_k\to g$ in measure and $|E|<\infty$.
We have this hint:
use the relation $f_kg_k-fg=(f_k-f)(g_k-g)+f(g_k-g)+g(f_k-f)$. Consider each term separately, using the fact that a function which is finite on $E, |E|<\infty$, is bounded outside a subset of $E$ with small measure.
What does the last sentence mean? I think $f$ and $g$ are only bounded a.e. on $E$. My approach to prove $fg_k \to fg$ in measure is defining $f_0 = \sup_{x\in E-Z} f(x)$ where $Z=\{x\in E:f(x) = \infty\}$. Then, we also have the relationship $$\left\{x\in (E-Z)- Q : |g_k-g|>\frac{\epsilon}{|f|} \right\}\subset \left\{x\in (E-Z)- Q : |g_k-g|>\frac{\epsilon}{|f_0|} \right\}, $$ where $Q=\{x\in E: f(x)=0\}$. The result follows if we take the measures of these two sets.
The definition of convergence I use is that $f$ and $\{f_k\}$ are measurable functions which are defined and finite a.e. in a set $E$. Then $f_k\to f$ in measure if for any $\epsilon>0$, $$\lim_{k\to\infty}m(\{|f-f_k|>\epsilon\}) = 0.$$
$\mu \{|f_kg_k-fg| >\epsilon\}$ $$ \leq \mu \{|(f_k-f)(g_k-g)| >\epsilon /3\}$$ $$+\mu \{|f(g_k-g)| >\epsilon /3\}+\mu \{|g(f_k-f)| >\epsilon /3\}.$$ Let us show that each of the three terms tend to $0.$ First term is $$\leq \mu \{|(f_k-f)| >\sqrt {\epsilon /3}\}+\mu \{|(g_k-g)| >\sqrt {\epsilon /3 }\} \to 0.$$ Let $M>0$ and consider $$\mu \{|f(g_k-g)| >\epsilon /3\} \leq \mu \{|(g_k-g)| >\epsilon /3M\}+\mu \{|f|>M\}.$$ Note that the sets $\{|f| >M\}$ decrease to empty set as $M \to \infty$. So we can choose $M$ such that $\mu \{|f|>M\}$ is as small as we want. [This requires the hypothesis $\mu (E) <\infty$]. Now it is clear that second term tends to $0$ and the third term is handled the same way.