Prove that $\lim \limits_{k\rightarrow \infty} \int \limits_E f_k g = \int \limits_E fg$

Question

I want to prove that $$\lim \limits_{k\rightarrow \infty} \int \limits_E f_k g = \int \limits_E fg$$ using Holders inequality. Assume that all of the conditions for Holders inequality are met ($f,g$ are Lebesgue measurable, $E$ is Lebesgue measurable, etc). I know that the norm is defined as $$ \parallel f \parallel_{p,E} = \left( \int \limits_E |f|^p \right)^{1/p} $$ So what I have is $$ \parallel f_k g \parallel_{1,E} \leq \parallel f_k \parallel_{p,E} \parallel g \parallel_{q,E} $$ Then I can take the limit of both sides $$ \lim \limits_{k\rightarrow \infty} \parallel f_k g \parallel_{1,E} \leq \lim \limits_{k\rightarrow \infty} \parallel f_k \parallel_{p,E} \parallel g \parallel_{q,E} = \parallel g \parallel_{q,E} \lim \limits_{k\rightarrow \infty} \parallel f_k \parallel_{p,E} $$ I shown before that $\lim \limits_{k\rightarrow \infty} \parallel f_k \parallel_{p,E} = \parallel f \parallel_{p,E} $ So then I have $$ \lim \limits_{k\rightarrow \infty} \parallel f_k g \parallel_{1,E} \leq \parallel f \parallel_{p,E} \parallel g \parallel_{q,E} $$ I can also say that $$ \parallel f g \parallel_{1,E} \leq \parallel f \parallel_{p,E} \parallel g \parallel_{q,E} $$ but I don't know if I am on the right track or even how to continue. Any ideas?

Answer 1

Assuming that your hypothesis is that $f_k\to f$ in $L^p$ (you don't say), $$ \left|\int_Ef_kg-\int_Efg\right|=\left|\int_E(f_k-f)g\right|\leq\int_E|f_k-f|\,|g|\leq \|f_k-f\|_p\,\|g\|_q \to0, $$ so $$ \lim_k\int_Ef_kg=\int_Efg. $$