Let $f_n$, $f \in L^2(0, 1)$ be such that $f_n$ converges to $f$ weakly in $L^2(0, 1)$. Let $g_n, g : (0, 1) \to \mathbb{R}$ be measurable functions such that $g_n$ converges to $g$ in measure and $\|g_n\|_{L^\infty} \leq M$ for every $n$. Prove that
a) $g\in L^\infty(0,1)$
b) $f_ng_n$ converges to $fg$ weakly in $L^2(0, 1)$.
My attempt: I think that point a) it's ok but i don't know how to finish point b).
a) Since $g_n$ converges to $g$ in measure for all $\varepsilon>0$ we have $$\lim_{n\to\infty} \mu(\{|g-g_n|<\varepsilon\})=1$$ We define $A^\varepsilon_n:=\{|g-g_n|<\varepsilon\}$ and $A^\varepsilon:=\bigcup_n A^\varepsilon_n$. By definition $A^\varepsilon_n \subset A^\varepsilon$ for every $n$, then $\mu(A^\varepsilon_n) \leq \mu(A^\varepsilon)$ and, taking the limit for $n\to +\infty$, we get $\mu(A^\varepsilon)=1$. Defined $$A:=\{\,x\in (0,1)\,|\,|g_n(x)|\leq M\,\},$$ it follows $\mu(A)=1$ (and so $\mu(A^\varepsilon \cap A)=1$). I claim that there exists $C>0$ such that $|g(x)|\leq C$ for every $x\in A^\varepsilon \cap A$. Indeed for every $x\in A^\varepsilon \cap A$, there exists $\overline n=\overline n_x$ s.t. $x\in A^\varepsilon_{\overline n_x}$ and so $$|g(x)|\leq |g(x)-g_{\overline n_x}(x)|+|g_{\overline n_x}(x)|< \varepsilon + M.$$ In conclusion, we have proved $|g(x)|< \varepsilon + M$ a.e., that is $||g||_{L^\infty}\leq M+\varepsilon.$
b) Let $h$ be an arbitrary function in $L^2$, then $$ \bigg|\int (f_n g_n - f g) h \, dx \bigg| \leq \bigg| \int (f_n g_n - f_n g)h \, dx\bigg| + \bigg|\int (f_ng - fg) h \, dx\bigg| =$$ $$= \bigg|\int f_n(g_n - g)h \, dx \bigg| + \bigg|\int (f_n - f)g h \, dx\bigg| $$ $$\leq (2M +\varepsilon)\bigg|\int f_n h \, dx \bigg| + \bigg|\int (f_n - f)g h \, dx\bigg| $$ And now ? The second integral goes to zero (since $gh\in L^2$ and $f_n \rightharpoonup f$), but what about the first one ? I know that convergence in measure implies other modes of convergence, but this only applies to subsequences.
Thanks in advance.