# $0\leq \int_{\mathbb{R}}fdm < \infty \implies m(E) = 0$?

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.

## 1 Answer

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 $$n >\frac {\int f dm} {m(E)}$$.

• Ah, of course! Thank you, Kavi. I must have been staring at this for too long to not have seen that. – Taylor Rendon Mar 3 at 23:59
• 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 $m(E) = 0$? – Taylor Rendon Mar 9 at 20:30
• @TaylorRendon $\int_{-1}^{1}x dx=0$ but $m(-1,1)=2$. – Kavi Rama Murthy Mar 9 at 23:09