Prove by contradiction that $|f(x)| < \infty$ almost everywhere for $f \in L^p$ ($p \ge 1)$

Question

Let $f \in L^p$, where $p \ge 1$, then $|f(x)| < \infty$ almost everywhere.

Does anyone know how to prove this by contradition?

Answer 1

A direct proof, from which a proof by contradiction can be derive if necessary.

Let $A_n=\{x \in \mathbb R \ : \ \vert f(x) \vert^p \ge n\}$ for $n$ integer and $b = \int \vert f(x) \vert^p dx$.

We have $$0 \le n \mu(A_n) \le b \tag{1}$$ where $\mu$ is the measure. Hence $\mu(A_n) \le \frac{b}{n}$ for all $n \in \mathbb N$.

$$C=\{x \in \mathbb R \ : \ \vert f(x) \vert = +\infty\} = \bigcap_{n=1}^\infty A_n. \tag{2}$$ Finally $\mu(C)=0$.

Now deriving a proof by contradiction from above.

$(A_n)$ is a decreasing sequence (for the inclusion order) of sets. If $\mu(C) > 0$, the equality $(2)$ implies that it exists $\delta > 0$ such that $\mu(A_n) \ge \delta$ for all $n \in \mathbb N$. Consequently $n \mu(A_n) \ge n \delta$ tends to $+\infty$ in contradiction with the inequality $(1)$.