I am working through Baby Rudin, and encountered the following remark:
If $f$ and $g\in\mathcal{L}(\mu)$ on $E$, and if $f(x)\leq g(x)$ for $x\in E$, then $$\int_{E}fd\mu\leq \int_E gd\mu.$$
Given a measurable function $f$, $\int_{E}fd\mu$ is always defined, but $f$ only belongs to $\mathcal{L}(\mu)$ when $$\Bigg|\int_{E}fd\mu\Bigg|<\infty.$$ (Edit: as harfe pointed out, it should be $\int_{E}|f|d\mu$ instead).
My question is, does the remark remain true if we only assume $f$ and $g$ to be measurable instead of being in $\mathcal{L}(\mu)$? It seems to be true, because of two cases:
- If $\int_{E}fd\mu=-\infty$, then there is nothing to show.
- If $\int_{E}fd\mu=\infty$, then it must follow (?) that $\int_{E}gd\mu=\infty$ as well, so we get $\infty\leq \infty$.
Any help appreciated.
Edit: I just realized that in my situation, I only need $f$ measurable and $g\in \mathcal{L}(\mu)$, but comments at any level of generality are welcome.
First, your definition of $\mathcal L(\mu)$ is not correct in my opinion. It is usually defined as the set of measurable functions $f$ such that $$ \int_E |f|\,\mathrm d\mu<\infty. $$
Your argument for the cases $\int_E f\,\mathrm d\mu=-\infty$ and $\int_E f\,\mathrm d\mu=\infty$ is correct. However, there is another possibility if $f$ is not in $\mathcal L(\mu)$. It can happen that the integral is not defined or does not exist. This is the case if $$ \int_E f^+\,\mathrm d\mu=\infty \quad\text{and}\quad \int_E f^-\,\mathrm d\mu=\infty, $$ where $f^+,f^-$ are the positive and negative part of $f$ (using the Notation as in Baby Rudin).
This is the reason why the author adds the conditions $f,g\in\mathcal L(\mu)$.