Lebesgue integrability implies finite almost everywhere

Question

Let $(X,\mathcal{M},\mu)$ be any positive measure space. Suppose $f \in \mathcal{L}(X,\mathcal{M},\mu)$. Prove that $f(x)$ must be finite $\mu$-almost everywhere.

I have defined a set $$E_n= [{x\in X : |f(x)|\geq n}]$$ which is a measurable set, and have that $$ \int_{E_n}{|f|}d\mu \leq \int_{X}{|f|}d\mu = C$$ for some constant C. At some point I'll have to take the limit of $n$ as $n$ approaches $\infty$, while at the same time using the fact that $E_\infty \subset E_n$. I am unsure of the details or whether this approach is correct.

Answer 1

Consider $n \chi _{E_n}$. Now for all $x$ in $E_n$ we have that $n\chi_{E_n}(x)=n\le |f(x)|$, using that the integral is monotone, we have

$$n \mu (E_n)=\int_X n\chi_{E_n} d\mu\le \int_{E_n} |f| d\mu\le \int_X|f|d\mu=C, $$

so, $\mu (E_n) \le C/n$. Set $E=\bigcap E_n$, i.e., $x$ belongs to $E$ iff $|f(x)|=\infty$. Then

$$0\le \mu(E)\le \mu(E_n )\le C/n,$$

letting $n\to \infty$, $\mu(E)=0$, that is, $f$ is finite $\mu$-a.e.

Answer 2

Suppose there is a set $E \subseteq X$ such that $\mu(E) > 0$ and $ f(x) = \infty$ for all $x \in X$. From the properties of the integral we have that \begin{equation*} \int_E f d\mu \leq \int_X f d\mu. \end{equation*} However, we also have that \begin{equation*} \int_E f d\mu \geq \inf_{x \in E}\left[f(x)\right] \mu(E) = \infty. \end{equation*}