If $f\in L^1(\mu)$ then $f$ is finite $\mu$-a.e.

Question

Prove that if $f,g\in L^1(\mu)$ then $f$ is finite $\mu$-a.e. (almost everywhere).

My proof:

Let $A=\{x:f(x)\geqslant n,\forall n\in\mathbb{Z}\}$

I intend to show $\mu(A)=0$

As $f\in L^{1}(\mu)$

$\int |f|d\mu\leqslant \infty$ then $\exists C$ such that $\int |f|d\mu<C$

Applying Chebyshev inequality:

$\mu(A)\leqslant\frac{1}{n}\int |f|d\mu<\frac{1}{n}C\implies \mu(A)<\frac{1}{n}C\\\implies\mu(A)<\frac{1}{n}C\implies\lim_{n\to\infty}\mu(A)<\lim_{n\to\infty}\frac{1}{n}C=0\implies \mu(A)=0 $

Questions:

Is my proof right? If not why not? What are the alternatives?

Thanks in advance!

Answer 1

Nice proof—it looks right to me.

What are the alternatives?

Personally I would write the proof a bit simpler. Let $A = \{x : f(x) = \infty\}$ (equivalent to your definition, but I'm not using $n$ and then taking a limit!) Now note that $A$ is a measurable set, so $$ \int |f| \, d\mu \ge \int_A |f|\, d\mu = \int_A \infty \, d\mu = \infty \cdot \mu(A). $$ Recall that in measure theory we define $0 \cdot \infty = 0$. Anyway, since $f$ is $L^1$, this integral should be finite, so we have to have $\mu(A) = 0$.