Let $([0,1],\mu)$ be a measure space. Let $f_n :[0,1] \to [0,\infty)$ be integrable functions. If $\int_{0}^{1} f_n d\mu = 1$ and $\int_{0}^{1/n} f_n d\mu > 1-1/n$ then show $\int_{0}^{1}\sup{f_n} d\mu=\infty$.
This seems to be true because $\sup f_n$ has a arbitrarily large value near $0$. But I have no idea how to handle such value has a enough width so that $\int_{0}^{1}\sup{f_n} d\mu=\infty$.
Is this a valid solution?
Note $g=\sup f_n$ dominates $f_n$. We have $\int_{1/n}^{1}f_n d\mu < 1/n$. Fix $N$ so on $[1/N,1]$ we have $\int_{1/N}^{1}f_n d\mu < 1/n$. Let $n\to \infty$. Then $f_n \to 0$ on $[1/N,1]$. Claiming this for all $N \in \mathbb{N}$ gets $f_n \to 0$ a.e. Now suppose $\int g d\mu <\infty$. Then applying DCT on $f_n$ gets contradiction. Because $\int f_n =1$ for all $n$ while $\int \lim f_n =0$.