Set $\{x | f(x) = \infty\}$ is of measure 0.

Question

Suppose we have a measurable function $f$ and $ \int f < \infty$ . We want to show that $\{x | f(x) = \infty\}$ is of measure 0. My attempt was the following: we can define the set $A = \{x | f(x) = \infty \}$ and we can define simple functions $\phi_n = n \chi_A$ where $\chi$ is the indicator function. Then, we see that $\phi_n \leq f$ $\forall n \geq 1$ and so we have that $\int \phi_n \leq \int f$ which implies that $n\chi_A \leq \int f$. From this point I don't know what to do or how to proceed. Is the process above a good one? How could I continue or can anyone give me another hint on how to prove this? Thanks!

Answer 1

Continuing your proof: ($\mu$ is a measure)

$n\mu(A) \leq \int f$ and $ \int f < \infty$. Let $ \int f =a$. Then, $\forall n$ we have, $n \mu(A) \leq a$ . Hence, $\mu(A) \leq a/n$. So, letting $n$ tends to $\infty$, we get, $\mu(A)=0$.

Answer 2

You've likely already shown the basic properties of integrals. Specifically $f$ is integrable iff $|f|$ is integrable, and that integrals are linear. So a proof by contradiction would be:

$$\int |f|d\mu=\int_{|f|=\infty}|f|+\int_{|f|<\infty}|f|\geq \infty$$.