Show that there exist $\gamma\in[\alpha,\beta] $ such that $\int\limits_{E}f|g|=\gamma\int\limits_{E}|g|$ 
Suppose $f:E\rightarrow \mathbb{R}$ is measurable, $g:E\rightarrow R$
  is Lebesgue integrable and there exists $\alpha,\beta\in\mathbb{R}$ such that
  $\alpha\leq f(x)\leq\beta$ for almost every $x\in E$. Show that there
  exist $\gamma\in[\alpha,\beta] $ such that $\int\limits_{E}
 f|g|=\gamma\int\limits_{E}|g|$ 

For this I can multiply the inequality by $|g|$ and get
$\alpha|g|\leq f|g|\leq\beta|g|$ 
Now IF I could apply the integration throughout the inequality then
$\int\limits_E\alpha|g|\leq \int\limits_Ef|g|\leq\int\limits_E\beta|g|$ 
So by taking $\gamma=\frac{\int\limits_Ef|g|}{\int\limits_E|g|}$ we can get the answer.  
But my doubt is whether it is possible to apply the integration mark along the inequality. Because the monotonicity of the Lebesgue integral was defined for non negative functions (although it was not integrable.) And I don't see a method to say that the function $f|g|$ is integrable.  
In my reference (that is Real analysis by Fitzpatrick) Lebesgue integrability for a measurable function were defined to be the case when: 
$$\int\limits_E|f|<\infty$$ 

Just to clarify that if $\displaystyle\int|g|=0$, then $g=0$ a.e. and $\gamma\in[\alpha,\beta]$ can be taken as arbitrary.
If it were not, then $\gamma=\dfrac{\displaystyle\int f|g|}{\displaystyle\int|g|}$ would be a candidate. And we know that $(\cdot)\geq(\cdot\cdot)$ implies that $\displaystyle\int(\cdot)\geq\int(\cdot\cdot)$ no matter $(\cdot),(\cdot\cdot)$ are positive or negative or not.
