If $f_n\rightarrow0$ and $|f_n|\leq\frac{1}{x\log x}$, prove that $\int_2^\infty f_ndx\rightarrow0$

Question

Let $g(x)=\frac{1}{x\log x}$ for $x>1$. For given measurable subsets of $[2,\infty]$ and non-negative sequence $c_n$, define $f_n(x):=c_n\chi_{A_n}(x)$. Prove or disprove that if $f_n\rightarrow0$ and $|f_n|\leq g(x)$, then $\displaystyle\int_2^\infty f_nd\mu_L\rightarrow0$. Note that $g$ is not integrable.

After so many examples, I believe it to be TRUE. After some trying to prove my guess, I reduced the problem to the case when $A_n=[a_n,b_n]$ and $\displaystyle\beta:=\sup_n b_n<\infty$.

Now I just need to prove $$c_n\mu(A_n)=c_n(b_n-a_n)\xrightarrow{n\rightarrow\infty}0$$

I'm mixed up a little here.

Answer 1

As said in comments, this is true.

1. Let $\beta_n=A_n$. Since $c_n\le \frac{1}{\beta_n\log \beta_n}$, it follows that $\int f_n \le 1/\log \beta_n$.

2. Thus, if the integrals $\int f_n$ fail to converge to zero, we can extract a subsequence with bounded $\beta_n$.

3. But as long as $\beta_n$ stays bounded, we have a uniformly bounded family of functions on a set of finite measure, so dominated convergence theorem applies.