Does a sequence bounded by a function of $L^{\infty}$, converges in $L^{\infty}$? I want to know if this statement is right?

Let $(g_n)$ be a measurable sequence from $[0,T]$ to $\mathbb{R}^n$
such that $\exists \beta \in L^{\infty}_{\mathbb{R}_+}([0,T])$ where
$|g_n(t)| \leq \beta(t) \forall t \in [0,T]$
then $(g_n) \in
 L^{\infty}_{\mathbb{R}^n}([0,T])$ and it admits a subsequence converges weakly*
to some $g \in L^{\infty}_{\mathbb{R}^n}([0,T])$ satisfying $|g(t)|
 \leq \beta(t)  \; \forall t \in [0,T]$

I try to prove it this way :

*

*since the functions $\{g_n, \forall n\}$ are measurable also bounded by $\beta$ then $\exists c >0$ such that $|g_n(t)| \leq \beta(t) \leq c$ then $g_n \in L^{\infty}_{\mathbb{R}^n}([0,T])$
2)$|g_n(t)| \leq \beta(t) \Rightarrow  | \frac{g_n(t)}{\beta(t)} | \leq 1 \Rightarrow (\frac{g_n}{\beta})$ in the unit ball of $L^{\infty}$ which is weakly* compact, then we can extract a subsequence $(\frac{g_k}{\beta})$ converges to $(\frac{g}{\beta})$ in that ball where $|g(t) | \leq \beta(t) $
Please tell me if something is wrong and thank you very much.

There's no need to divide by $\beta$. Since $||g_n||_\infty$ is bounded there is a weak* convergent subsequence; let's save typing by saying $g_n\to g$ weak*.
Now fix $\lambda>0$ and let $E$ be the set where $\beta\le\lambda$. Then $g_n\chi_E\to g\chi_E$ weak* and $||g_n\chi_E||_\infty\le\lambda$, hence $||g\chi_E||_\infty\le\lambda$. So $|g|\le\beta$ almost everywhere.
