Showing that a measure is finite

Question

I try to solve the following task and I'm not sure, if I did it correct:

Let $(\Omega,\mathfrak{A},\mu)$ be a measure space and $f:\Omega\rightarrow]0,\infty[$ is a measurable function so that $f$ and $\frac{1}{f}$ is $\mu-$integrable. Prove that $\mu$ is a finite measure.

My thoughts:

$\mu$ finite means that $\int \chi_\Omega d\mu = \mu(\Omega)<\infty$. Since $f,\frac{1}{f}$ are integrable we know that: $$\int\limits_\Omega f(x)d\mu <\infty \ \ and \ \int\limits_\Omega \frac{1}{f(x)}d\mu <\infty$$

So $$\infty>\int \chi_{\Omega}*f(x) d\mu \int \chi_{\Omega}*\frac{1}{f(x)}d\mu = \int \chi_{\Omega}*f(x)\frac{1}{f(x)}=\int \chi_\Omega d\mu$$

Is this correct?

Answer 1

No it is not. In general we will have $$ \int_\Omega f(x)\, d\mu \cdot \int_\Omega g(x)\, d\mu \ne \int_\Omega f(x) g(x)\, d\mu $$ To prove your result, use $$ \chi_\Omega = \sqrt f \cdot \frac 1{\sqrt f} $$ and Hölder's inequality with $p = q = 2$.

Answer 2

Estimate: $\mu(f>1)\le\int_{\{f>1\}} f\,d\mu\le\int_\Omega f\,d\mu<\infty$ and $\mu(f\le 1)\le\int_{\{f\le1\}} {1\over f}\,d\mu\le\int_\Omega {1\over f}\,d\mu<\infty$. Therefore $\mu(\Omega)=\mu(f>1)+\mu(f\le 1)<\infty$.