Hello,
No you just have to say that , the limit is 1 if x=0. So that's almost everywhere in [0,1] (you just have to exclude 0, and since we're dealing with the Lebesgue measure, {0} is of measure 0)
That's a tad simpler
suppose we have a space , where is Lebesgue measure. and let be a sequence of functions:
Show in measure and almost eveywhere.
To show the convergence in measure, for given , let . Then e^{-x})^n > \epsilon \}=" alt="E=\{x:f_n>\epsilon \}=\{xe^{-x})^n > \epsilon \}=" /> . So as . So in measure.
here is my question. In this question, i dont really see the difference between the convergence in measure and convergence almost everywhere. Can i prove the convergence almost everywhere in the same way i did above? or i have to show the different proof?