given that：$g(x)=\frac{1}{xln(x)},x > 1$. $f_n=c_n\mathbf{1}_{A_n}$, where $c_n \geq 0$, $A_n$ is measurable and $\subset [2, +\infty)$. $|f_n|\leq g$ ($f_n$ is dominated by $g$.) $\forall x, \lim_{n \rightarrow \infty}f_n(x) = 0$.
the question is：do we have$\int^\infty_2 f_n(x)dx \rightarrow 0, n \rightarrow \infty$?
Yes.
The function $g$ is not Lebesgue integrable, since monotone convergence theorem shows the calculation below is valid. $$\begin{align} \int_2^\infty g(x)dx &= \lim_{R\to\infty}\int_2^R g(x)dx\\ &=\lim_{R\to\infty}[\log\log x]_2^R\\ &= \infty \end{align}$$
Then you can't use Lebesgue's dominated convergence theorem. However, for arbitrary sequence $(f_n=c_n\chi_{A_n})$ which is dominated by $g$ and $f_n\to 0$ pointwise, it can be said that $\int_2^\infty f_n(x)dx \to 0$. (Here, $\chi_A$ is a characteristic function of $A$.) We show this by indirect proof and assume $\int_2^\infty f_n(x)dx \not\to 0$.
First we claim $$\forall \varepsilon\in (0,1/2), \exists ! t\in(2,\infty), g(t)=\varepsilon$$ This is easy since $g$ is decreasing in $(2, \infty)$ and $g(2)>1/2, \lim g(x) = 0$. Now we can say that for arbitrary function $\varepsilon\chi_{A}$ dominated by $g$ and $1/2$, $$\int_2^\infty \varepsilon\chi_{A} \leq \varepsilon (t-2)\leq \varepsilon t=\frac{1}{\log t}$$
Note that if $\varepsilon \to 0$, then $t \to \infty$, then $\frac{1}{\log t}\to 0$.
Since $\int_2^\infty f_n(x)dx \not\to 0$, you can take a subsequence $(f_{n_k})$ such that $\int_2^\infty f_{n_k}(x)dx \to I$, where $I=\limsup_n \int_2^\infty f_n(x)dx \in (0, \infty)$.
Let $\varepsilon_0\in (0,1/2)$ be a real number such that $\frac{1}{\log t_0} < I/2$. ($t_0$ is a unique number satisfying $g(t_0)=\varepsilon_0$.) There exists a number $N$ such that for all $k>N$, $\int_2^\infty f_{n_k}(x)dx > I/2$. By the argument above, it can be said that $c_{n_k} > \varepsilon_0$.
Applying Lebesgue's dominated convergence theorem to $(f_{n_k})$ with dominating function $g\chi_{[2, t_0]}$, it follows that
$$\lim_{k\to \infty}\int_2^\infty f_{n_k}(x) dx = \int_2^\infty 0 dx = 0 $$
However this is a contradiction since $(f_{n_k})$ was taken to satisfy $\int_2^\infty f_{n_k}(x)dx \to I$.