# Thread: convergence in measure and convergence a.e

1. ## convergence in measure and convergence a.e

suppose we have a space $([0,1],B,\lambda )$, where $\lambda$ is Lebesgue measure. and let $f_n(x)=e^{-nx}$ be a sequence of functions: $[0,1]\rightarrow [0,1]$

Show $f_n \rightarrow 0$ in measure and almost eveywhere.

To show the convergence in measure, for given $\epsilon >0$, let $E=\{x:f_n>\epsilon\}$. Then $E=\{x:f_n>\epsilon \}=\{xe^{-x})^n > \epsilon \}=" alt="E=\{x:f_n>\epsilon \}=\{xe^{-x})^n > \epsilon \}=" /> $\{x:e^{-x}>\epsilon ^{1/n}\}=\{x<-Ln(\epsilon ^{1/n})\}=[0,-Ln(\epsilon)/n)$. So $\lambda ( [0,-Ln(\epsilon)/n)) \rightarrow 0$ as $n \rightarrow 0$. So $f_n \rightarrow 0$ 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?

2. Hello,

No you just have to say that $\lim_{n\to\infty} f_n(x)=0 ~,~ \forall x\in (0,1]$, 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)