Show that there is no Lebesgue integrable function $g$ satisfying $n\chi_{(0,\frac1n]}\leqslant g$ for all $n$.

Question

Show that there is no Lebesgue integrable function $g$ satisfying $n\chi_{(0,\frac1n]}\leqslant g$ for all $n$, where $\chi$ is the characteristic function.

I tried to reason by contradiction. Assume there is such $g$. Then, fix arbitrary $n$ and take $x=\frac1n$ so that $n\leqslant g(x)$. Since $n$ was random, we can let $n\to\infty$ and $g(x)=\infty$. But then $g$ cannot be Lebesgue integrable.

I sense that something is wrong with this. (The part in italics.) For one, by the same reasoning we can "show" that $\int n\chi_{(0,\frac1n])}\,d\lambda=\infty$, when in fact it is equal to $n\lambda((0,\frac1n])=1$.

(1) Can somebody please point out what exactly is wrong with my "proof"; I know it is wrong but I cannot see where it fails.

(2) Any thoughts about how to find a correct proof will be helpful.

Thanks in advance!

Answer 1

(1) You have $n\le g(x)$ when $x\in [0,\frac 1n]$. So the $x$'s are changing, you cannot take $n\to \infty$ while fixing $x$.

(2) Indeed you have $g(x)\ge n$ when $x\in [0,\frac 1n]$. In particular, $g(x) \ge n$ for $x\in (\frac {1}{n+1}, \frac 1n]$. So $g(x)\ge f(x)$, where $f(x) = n$ when $x\in (\frac{1}{n+1}, \frac 1n]$. Is this $f$ integtrable?

Answer 2

Let $g$ be such a Lebesgue integral function. Then $g|_{[0,1]}$ is Lebesgue integral and positive so that $0 < \int_0^1 g < + \infty$. Pick any positive integer $n \ge \int_0^1 g$: then we have $$n+1 = \sum_{k=1}^{n+1} 1 = \sum_{k=1}^{n+1} \int_{1/(k+1)}^{1/k} k \le \int_0^1 g \le n$$ a contradiction.

Answer 3

Let $s:=\sum_{n=1}^{\infty}\chi_{(0,\frac1{n}]}$ so that $\int sd\lambda=\sum_{n=1}^{\infty}\frac1{n}=+\infty$.

This with $s=\sum_{n=1}^{\infty}n\chi_{(\frac1{n+1},\frac1{n}]}\leq g$.

Consequently $\int gd\lambda=+\infty$.