I was able to simplify the equation
$g(a)=\frac{1}{ea^{r}}\Bigg(\int^{y_l}_{-\infty} \ln\left(\frac{1}{ea^{r}}\right)f_0(y)dy$
$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{1-\frac{\ln a}{\ln(ab)}}(y)\ln\left( \frac{1}{ea^{r}}e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{-\frac{\ln a}{\ln(ab)}}(y)\right)dy+\int^\infty_{y_u} a\ln\left( \frac{a}{ea^r}\right)f_0(y)dy \Bigg)=\epsilon $
using
$z=ea^{r}=\left(\int^{y_l}_{-\infty} f_0(y)dy+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{ln(ab)}}{f_1}^{\frac{lna}{ln(ab)}}(y){f_0}^{1-\frac{lna}{ln(ab)}}(y)dy+\int^\infty_{y_u}a f_0(y)dy \right) $
we have
$g(a)=\frac{1}{z}\Bigg(\int^{y_l}_{-\infty} -\ln\left(z\right)f_0(y)dy$
$+\int^{y_u}_{y_l}-\ln{z} e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{1-\frac{\ln a}{\ln(ab)}}(y)dy$
$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\ln\left( e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\right){f_0}(y)dy$
$+\int^\infty_{y_u} -\ln{z}\,af_0(y)dy+ \int^\infty_{y_u} a\ln af_0(y)dy \Bigg) $
collecting the terms with $-\ln z$ in common parenthesis we have
$g(a)=\frac{1}{z}\Bigg(-\ln\left(z\right)\Bigg(\int^{y_l}_{-\infty} f_0(y)dy$
$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{f_1}^{\frac{\ln a}{\ln(ab)}}(y){f_0}^{1-\frac{\ln a}{\ln(ab)}}(y)dy+\int^\infty_{y_u} \,af_0(y)dy\Bigg) $
$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\ln\left( e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\right){f_0}(y)dy$
$+ \int^\infty_{y_u} a\ln af_0(y)dy \Bigg) $
The inner parenthesis is also $z$ accordingly we can write
$g(a)=\frac{1}{z}\Bigg(-\ln\left(z\right)z $
$+\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\ln\left( e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\right){f_0}(y)dy$
$+ \int^\infty_{y_u} a\ln af_0(y)dy \Bigg) $
which is
$-\ln\left(z\right)+\frac{1}{z}\Bigg(\int^{y_u}_{y_l} e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\ln\left( e^{\frac{\ln^2a}{\ln(ab)}}{l}^{\frac{\ln a}{\ln(ab)}}(y)\right){f_0}(y)dy+ \int^\infty_{y_u} a\ln af_0(y)dy \Bigg)$
After this point, what I know is that, the derivative of $z$ with respect to $a$ is positive and the derivative of $l$ with respect to $y$ is also positive. What I need to show is that $g(a)$ is monotone! however the last expression that I derived depends on $y_l$ at the limits of the integral and I dont know how to deal with this case, especially since $y_l=l^{-1}(1/a)$