I am currently reading my textbook on measure theory and started reading the author's proof for the following lemma regarding the Lebesgue integral.

Let $f$ be a Lebesgue measurable function on $\mathbb{R}$. Then $f$ is finite almost everywhere.

I understand all of the first bit, which he writes as (in my own words):

It suffices to consider the case $f \geq 0$. Let $n \in \mathbb{N}$, and define $E_n:=\{x \in \mathbb{R}: f(x) \geq n\}$. Therefore take $E:= \bigcap_{n=1}^{\infty}E_n$ such that $$\int_{\mathbb{R}}fdm \geq \int_{E_n}fdm \geq n\cdot m(E_n) \geq n \cdot m(E),$$ and It follows that $m(E) \leq \frac{1}{n}\int_{\mathbb{R}}fdm \ \forall n \in \mathbb{N}$ . However, following this sentence is where I start to get lost. He states that "since $0\leq \int_{\mathbb{R}}fdm < \infty$ we can conclude that $m(E) = 0$". What is the reasoning behind this? For some reason, I can't seem to see why $0\leq \int_{\mathbb{R}}fdm < \infty$ implies that $E$ is null. I am pretty new to Lebesgue integration so if this is due to some basic fact, forgive me.

Any feedback is welcome.


In line 3 you should say 'Let $f$ be a Lebegue integrable function on $\mathbb R$'.

The inequality holds for all $n$. If $m(E)>0$ you get a contradiction by chosing an integer $n >\frac {\int f dm} {m(E)}$.

  • $\begingroup$ Ah, of course! Thank you, Kavi. I must have been staring at this for too long to not have seen that. $\endgroup$ – Taylor Rendon Mar 3 at 23:59
  • $\begingroup$ I have a quick follow up question: if we were given that $\int_{E}fdm = 0$ for some measurable set $E$, could we conclude that $m(E) =0$ by saying that since $m(E) \leq \frac{1}{n} \int_{E}fdm$? Since $\int_{E}fdm=0$ would force $m(E) = 0$? $\endgroup$ – Taylor Rendon Mar 9 at 20:30
  • 1
    $\begingroup$ @TaylorRendon $\int_{-1}^{1}x dx=0$ but $m(-1,1)=2$. $\endgroup$ – Kavi Rama Murthy Mar 9 at 23:09

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