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If $\int_0^\infty f(x) dx\leq\infty$ , we regard it as "convergent" , does there exist such an $f$, which satisfies $\int_0^\infty f(x) dx<\infty$ , but $\int_0^{\infty}k\chi_{E_k}(x)dx$ "divergent" ?

Where $E_k=\{x:f(x)>k\}$

Well , I came to this question when I was trying to prove this

If an improper integral of $f$ is finite, then $f<\infty$ for a.e.x.

We know that in the Lebesgue Integrals, one has if $f$ is Lebesgue integrable then $f<\infty$ for a.e.x.

I was trying to modifying the proof of that . Let $E_k=\{f>k\}$ , and $k\chi_{E_k}<f$ , then take the improper integral both side of the inequality , however , a problem comes , does $k\chi_{E_k}$ integrable ? Well, I just afraid of that may be a Dirichelet-like function , then that will come a "Divergent" case

Alexander Lau
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  • Is $f$ assumed to be non-negative? Also, I think the first $\leq$ should be a $<$. – carmichael561 Apr 15 '19 at 01:57
  • @carmichael561 Yes , when f is non-negative, then one can use the comparison principle , but $k\chi_{E_k}$ may be even nonintegrable , so I confused, I had refreshed the question. – Alexander Lau Apr 15 '19 at 02:17

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I think I got one , in the extended sense , if $f>0$ and integrable , then the improper integral exists,thus the problem is to ask if $f$ is integrable , whether $\chi_{\{f-k>0\}}$ integrable ? Well, since the $f-k$ is integrable , the question is

If $ f\geq0$ is integrable , whether $\chi_{\{f>0\}}$ is integrable ?

The answer is no, for example , the well known Riemann function $R(x)$, which is non-nagative integrable , but $\chi_{\{R>0\}}=D(x)$ which is not integrable.

However, it won't stop me proving that result , i.e:

the improper integral of $f$ is finite ,then $f$ is a.e finite.

For I can use $\{f\geq k\}$ to replace the original $E_k$

Alexander Lau
  • 129