Hi everyone I'm not completely familiar with this kind of argument and I'd appreciate if someone can help me to see if the argument is correct and also any suggestion to improve it. Thanks in advance.
Let $f_n=1_{[0,n]}/n^2$. Is there a function $g$ (for the Lebesgue measure on $\bf{R}$) dominating these $f_n$?
Let $g(x)=1_{[0,1)}(x)+\lfloor x^2 \rfloor^{-1} 1_{[1,\infty)}(x)$. We shall show that $g$ is measurable and dominates each of the $f_n$. Let $E_n= [n,n+1)$ for all $n\in \bf{N}_{\ge 1}$, as $g$ is measurable in $ (-\infty, 0)$, $ [0,1)$ and in each $ E_n$ with the relative sigma algebra and the union of each one is $\bf{R}$, therefore $g$ is measurable.
Let $x\in {\bf{R}}$ and $n \in \bf{N}_{\ge 1}$, the result is trivial when $n=1$, we may assume that $n>1$. If $x\in [0,1)$, then $f_n(x)=1/n^2< g(x)=1$, if $x\in [1,n]$, then $f_n(x)=1/n^2\le g(x)$ as is the smallest value which $g$ takes in this interval. When $x\in (-\infty, 0)\cup (n,\infty)$ the result is trivial. Then $f_n(x)\le g(x) $ for each $x\in \bf{R}$.
To conclude we prove that $g$ is absolutely integrable. We observe that if $x\in [n,n+1)$ for all $n\ge 1$, then $g(x)=1/n^2$. Let $F_N=\lfloor x^2 \rfloor^{-1} 1_{[1,N+1)}(x)$ for each $N\ge 1$ it's not difficult to see that $0 \le F_N(x)\uparrow\lfloor x^2 \rfloor^{-1} 1_{[1,\infty)}(x)$, so for monotone convergence theorem we have
$$\int_{\bf{R}} \lfloor x^2 \rfloor^{-1} 1_{[1,\infty)}(x) d\lambda = \lim_N \int_{\bf{R}} F_N d\lambda = \lim_N \int_{\bf{R}} \sum_{n\le N}1/n^2 1_{[n,n+1)} d\lambda= \sum_n^\infty 1/n^2< \infty$$
For what we have seen and using the linearity of the integral
$$\int_{\bf{R}}g d\lambda= \int_{\bf{R}}1_{[0,1)}(x) d\lambda + \int_{\bf{R}}\lfloor x^2 \rfloor^{-1} 1_{[1,\infty)}(x) d\lambda= 1+\sum_n 1/n^2<\infty$$
Therefore, $g$ dominates each of the $f_n$ as claimed.