# Show that $\int_0^1 \sup_{n\in\mathbb N}\{ f_n(x) \}dm(x)=\infty$

Let $$m$$ denote Lebesgue measure on $$[0,1]$$. Let $$\{f_n\}$$ be a sequence of Lebesgue measurable functions on $$[0,1]$$ with values in $$[0,\infty]$$ such that $$\lim_{n\to\infty} f_n(x)=0$$ almost everywhere and $$\int_0^1 f_n(x)dm(x)=1\quad\text{for all}\ n.$$ Define $$g(x)=\sup_{n\in\mathbb N}\{ f_n(x) \}$$. Show that $$\int_0^1 g(x)dm(x)=\infty.$$

My attempt:

Since $$f_n(x)$$ converges to $$0$$ almost everywhere and $$m([0,1])=1<\infty$$, by Egorov's theorem we know that for every $$\delta>0$$, $$f_n(x)$$ converges to $$0$$ uniformly on the complement of a measurable set $$E_{\delta}\subset [0,1]$$ with $$m(E_\delta)<\delta$$. Note that $$\int_0^1 f_n(x)dm(x)=1$$ for all $$n$$ which means $$1=\int_0^1 f_n(x)dm(x)=\int_{E_\delta}f_n(x)dm(x)+O\left(\frac 1n\right) .$$ I was about to show that ''most'' of the area under $$f_n$$ is counted towards the integral of $$g(x)$$ and apply the Egorov's theorem again to $$E_\delta$$ but I failed.

• If $g$ were integrable you could apply the dominated convergence and switch limits and integrals – Conrad May 17 '19 at 10:34

Suppose that $$\int_0^1 g(x)dm(x)<\infty.$$ Since $$g \ge 0$$ and $$g$$ is measurable, $$g$$ is integrable.
Furthermore we have $$0 \le f_n \le g$$ for all $$n$$ on $$[0,1]$$.