Let $(X, \Sigma, \mu )$ be a measure space. If $f: X \rightarrow \bar{\mathbb R}$ is measurable and $\int |f| \, d \mu < \infty$, then:
for any $a>0$, the set $\{x \in X : |f(x)|>a \}$ has finite measure
the set $\{x \in X: f(x) \neq 0\}$ has $\sigma$-finite measure, i.e. is a countable union of sets of finite measure.
I am really trying to learn this myself and I would love to see how these two statements would be proved.
Please can someone help me.
On
Ok, Nigel got part $1$. For part $2$,
$\{x \mid f(x) \neq 0\} = \{ x \mid |f(x)| \neq 0 \} = \bigcup \limits_{n = 1}^{\infty} \{x \mid |f(x)| > \frac{1}{n} \}$. Of course, you should prove that these two sets are equal, but once you do, each of the sets in the union has finite measure by part $1$, and thus the set $\{ x \mid f(x) \neq 0 \}$ is $\sigma$-finite.
$1)$ Assume that the set $A = \{ x \in X : |f(x)| > a \}$ has infinite measure, then we have $$\int_X|f| \mathrm{d}\mu = \int_A|f|\mathrm{d}\mu + \int_{A^c}|f|\mathrm{d}\mu \geq \int_A |f|\mathrm{d}\mu \geq a\cdot\mu(A) = \infty$$ which is contradiction to our hypothesis that $f$ is integrable.
$2)$ See the answer by user46944.