C is a Countable set. $g$ is continuous in $ [1,\infty)$ and $\phi(x)>0, \forall x\in[1,\infty)-C$. $\phi(x)$ and $g(x)\phi(x)$ are integrable in $[1,\infty)-C$.
Prove that there exists a $c\in(1,\infty)-C$ such that:
$$\frac{\displaystyle\int_{(1,\infty)-C} g(x)\phi(x) dx}{\displaystyle \int_{(1,\infty)-C} \phi(x)dx }=g(c)$$ g(x)= $(x^{1-\sigma}-x^\sigma)cos (t lnx), 0<\sigma<1$ $\phi(x)=1-x+[x]/x^2$ where [.] denotes the greatest integer function. $C= \{x \mid cos (tlnx)=0\}$
You can't prove it, because it is not true.
Consider $\phi(x)=g(x)=e^{-x}$ and $C=\{1+\log 2\}$. Then the left-hand-side is $1/2e$, but $g(x)=1/2e$ if and only if $x=1+\log 2$.
(Note that if you drop the condition $c\notin C$ and just require that $c\in(1,\infty)$, then the usual proof of the mean value theorem for integral works.)