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$$m(\{x \in E \;|\; f(x) \geq \lambda \} )= m(A) \leq \frac{1}{\lambda}\int_E f $$$$=\frac{1}{4}\int_0^1\frac{\ln t\ln^2(1-t)}{\sqrt t\sqrt{1-t}}dt=\frac{1}{4}\frac{d^2}{db^2}\frac{d}{da}\int_0^1t^{a-1/2}(1-t)^{b-1/2}dt|_{a=b=0}$$$$m(\{x \in E \;|\; f(x) \geq \lambda \} )= m(A) \leq \frac{1}{\lambda}\int_E f $$$$=\frac{1}{4}\int_0^1\frac{\ln t\ln^2(1-t)}{\sqrt t\sqrt{1-t}}dt=\frac{1}{4}\frac{d^2}{db^2}\frac{d}{da}\int_0^1t^{a-1/2}(1-t)^{b-1/2}dt|_{a=b=0}$$m({x in E : f(x) >= \lambda})= m(A) <= (1/\lambda)*\int_E f= 1/4*\int_0^1[(ln(t)*ln^2(1-t))/(t*(1-t))^.5] dt = 1/4*(d^2)/(db^2)*(d)/(da)*\int_0^1[t^{a-1/2}(1-t)^{b-1/2}] dt|_{a=b=0}Ads