I have been going through a course in measure theory and integration. Got stuck in one intermediate step of the proof.It says that-
If $g$ is integrable on $\mathbb{R}$ and $f$ is measurable and there exist $\alpha, \beta$ such that $\alpha\le f(x)\le\beta$
Then there exists $\gamma\in [\alpha, \beta]$ such that
$$\int f \lvert g\rvert dx=\gamma \int \lvert g\rvert~dx$$
This has been my attempt.
Since $\alpha$ $\le$ $f$ $\le$ $\beta$
Therefore $\lvert f\rvert$ $\le$ $\delta$ where $\delta$ = max{$\lvert \alpha\rvert$, $\lvert \beta\rvert$}
Therefore $\lvert f\rvert$$\lvert g\rvert$ $\le$ $\delta$$\lvert g\rvert$
From here can I conclude that $f$ $\lvert g\rvert$ is integrable? (This is my main doubt)
Best Answer
This follows from the comparison test for Lebesgue integrability or proved as follows.
Let $\mathcal{H}$ be the set of all nonnegative, bounded measurable functions of finite support such that $0 \leqslant h \leqslant |f||g|$. Since $|f||g| \leqslant \delta|g|$ and $\delta |g|$ is integrable we have by monotonicity of the integral,
$$\int h \leqslant \int \delta |g| \implies \int|f||g| = \sup_{h \in \mathcal{H}}\int h \leqslant \int \delta |g| < +\infty,$$
and, thus, $|f||g|$ is integrable. Since $f^+|g|, f^-|g| \leqslant |f||g|$ it follows that $f|g| = (f^+-f^-)|g|$ is integrable.