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If I assume $f$ is any real measurable function and $g$ integrable function. Now, let $\alpha, \beta \in R$ such that $\alpha \leq f \leq \beta $ $a.e.$. I want to prove that there exists $\gamma$ with $\alpha \leq \gamma \leq \beta$ such that $\int f|g|dx = \gamma\int |g|dx$. How I do this? Thank you!

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2 Answers 2

Hint: intermediate value theorem.

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I try to do this by using Intermediate Value Theorem, however I have problem, that is the Lebesgue integral need not be continuous. – Nimana Jan 28 '13 at 12:35
$\gamma \int |g|\ dx$ is a continuous function of $\gamma$. – Robert Israel Jan 28 '13 at 19:22

Suppose $g$ is not zero. Set $$ \gamma = \frac{\int f|g|}{\int |g|}.$$

Then prove $\alpha \leq \gamma \leq \beta$.

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I think you wanted to say "suppose $\int \lvert g \rvert \neq 0$". – epimorphic Jan 12 at 0:56

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