I already saw some proofs here with $b_n\to b$ in $L^2$ and $a_n\rightharpoonup a$ in $L^2$. Then $$ \int a_n b_n \to \int a b. $$
But what goes wrong if both sequences are weak convergent?
Proof:
$$ \int_\Omega a_nb_n\,dx \to \int_\Omega ab\,dx. $$ Indeed, $$ a_nb_n-ab=a_n(b_n-b)+(a_n-a)b $$ Then $$ \Big|\int_\Omega a_n(b_n-b)\,dx\,\Big|\le \|a_n\|_{L^2}\|b_n-b\|_{L^2}\le M\|b_n-b\|_{L^2}\to 0, $$ and $$ \int_\Omega (a_n-a)b \to 0, $$ as $a_n-a\rightharpoonup 0$.
Why does not hold $$ \int_\Omega a_n(b_n-b)\,dx\,\to 0,$$ if $b_n\rightharpoonup b$.
Then it may fail. Let $a_n(x)=b_n(x)=\sin(n\,x)$. By the Riemamm-Lebesgue lemma, $a_n$ and $b_n$ converge weakly to $0$ on $[-\pi,\pi]$, but $$ a_n\,b_n=\sin^2(x)=\frac{1-\cos(2\,n\,x)}{2} $$ converges wekly to $1/2$.