Does $f_n\to{}0$ and $\int_{[0,1]}f_nd\mu=1$ for every $n\geq{}1$ imply $\int_{[0,1]}\displaystyle\sup_{n}f_nd\mu=\infty$?

Question

Let $f_n:[0,1]\to\mathbb{R}^+$ such that $f_n\to{}0$ pointwise and $\int_{[0,1]}f_nd\mu=1$ for every $n\geq{}1$.

Does this imply $\int_{[0,1]}\displaystyle\sup_{n}f_nd\mu=\infty$?

Considering the most common example $f_n(x)=n\mathbb{1}_{[0,\frac{1}{n}]}(x)$ we have $\displaystyle\sup_{n}f_n(x)=n \iff x\in{}(\frac{1}{n+1},\frac{1}{n}]$ and thus $\int_{[0,1]}\displaystyle\sup_{n}f_nd\mu=\sum_{n=1}^{\infty}\int_{\frac{1}{n+1}}^{\frac{1}{n}}\displaystyle\sup_{n}f_nd\mu=\sum_{n=1}^{\infty}\frac{n}{n(n+1)}=\sum_{n=1}^{\infty}\frac{1}{n+1}=\infty$ which suggests it might be true.

Is the result true in general?

Answer 1

Yes. If $\int\sup_n f_n<\infty$ then you could let $g=\sup_n f_n$ and use dominated convergence to derive a contradiction.

Answer 2

Yes, it is. If $\sup_n f_n\in L^1([0,1]),$ then $\lim_{n\to \infty} \int_0^1 f_n\textrm{d}\mu=\int_0^1 \lim_{n\to\infty} f_n\textrm{d}\mu=0$ by the dominated convergence theorem.