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.

  • $\begingroup$ If $g$ were integrable you could apply the dominated convergence and switch limits and integrals $\endgroup$ – 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 $[0,1]$.

Now invoke the dominated convergence theorem to get a contradiction.

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