As I continue going through measure theory with Folland and the introductory text by Tao, I came across this simple problem in a problem note online regarding unsigned Lebesgue integrals.
Let $(X, F, \mu)$ be a measure space and $f \in L^+ (X, F)$ such that $\int f \,d\mu < \infty.$ Show that $\{f=\infty\} = f^{-1} (\infty)$ is a null set.
Here $L^+$ is the set of non-negative measurable functions while the integral $\int fd\mu$ is the supremum of simple functions less than or equal to $f$.
Now I tried to prove this by contradiction. First setting $\{f=\infty\}$ to be a set $K$ and supposing $\mu(K) > 0$, then we can define some $g_n = n\cdot \textbf{1}_K$. Clearly $g_n \leq f$ and $g_n$'s are simple for any $n > 0$.
Hence $\int g_n d\mu = n\cdot\mu(K) \leq \int f d\mu$. Finally taking $n \to \infty$ we would have the integral $\infty \cdot \mu(K) = \infty \leq \int f d\mu < \infty$ (given) and we are done.
Does this approach seem feasible under the given constraints for simple functions and Lebesgue integrals?
Best Answer
Yes, as you stated, if $\mu\{f=\infty\}>0$, then
\begin{align} \infty=\infty\cdot\mu\{f=\infty\}=\lim_{n\to\infty}\int g_nd\mu\leq\int_{\{f=\infty\}} fd\mu \leq \int fd\mu<\infty. \end{align} Hence, $\int fd\mu=\infty$ when $\mu\{f=\infty\}>0$ and we have a contradiction.