Let $\Omega =(0,1)$ equipped with the Borel sets and Lebesgue measure. Let $\alpha\in (1,2)$ and $X_n=n^\alpha \chi_{(1/(n+1), 1/n)}\rightarrow 0$ a.s. Show that Theorem 1.6.8 can be applied with $h(x)=x$ and $g(x)=|x|^{2/\alpha}$, but the $X_n$ are not dominated by an integrable function.
I did not understand the bold part i.e. "but the $X_n$ are not dominated by an integrable function" and why $X_n=n^\alpha \chi_{(1/(n+1), 1/n)}\rightarrow 0$ a.s.. Surely $n^{\alpha}\to\infty$ as $n\to\infty$. But $\chi_{(1/(n+1), 1/n)}\rightarrow $ zero measure set. So $\lim_{n\to\infty} X_n$ is like a infinity stick on the point $0$.
In rigorous process:
$$\sum_{n=1}^{\infty} P(|X_n|>\epsilon)< \sum_{n=1}^{\infty} \frac{EX_n^2}{\epsilon^2}$$ which is clearly not a finite quantity. So I could not come to a conclusion.
If we put this problem aside what does but the $X_n$ are not dominated by an integrable function means?
Theorem 1.6.8: Suppose $X_n\rightarrow X$ a.s.. Let $g, h$ be continuous functions with
(i) $g\geq 0$ and $g(x)\to\infty$ as $|x|\to\infty$.
(ii) $|h(x)|/g(x)\to 0$ as $|x|\to\infty$.
and (iii) $Eg(X_n)\leq K <\infty$ for all $n$.
Then $Eh(X_n)\to Eh(X)$.
Again, I do not need the proof of the application of the theorem, I need to know what means only the bold part.