0
$\begingroup$

Let's consider a sequence of functions $f_n:[a,b] \rightarrow \mathbb{R}, f_n(x)=e^{-n|1-sin(x)|}$. Show that $f_n$ converges to $f=0$ in measure.

Attempt/Thoughts: To prove $f_n$ converges to $f=0$ in measure, I have to show that for every $\epsilon >0$, $\mu (x:|e^{-n|1-sin(x)|}| \geq \epsilon) \rightarrow 0$ as $n \rightarrow \infty$. This is clear to me pictorially as seen when we graph the sequence of functions, as n gets larger and larger, the set of $x$ for which $f_n(x)$ is positive grows smaller and smaller. The peak grows narrower and narrower. Thus for me, it is clear pictorially that $\mu (x:|f_n(x)-f(x)| \geq \epsilon) \rightarrow 0$ as $n \rightarrow \infty$. enter image description hereexactly sure how to prove this formally, any

However, I'm not exactly sure how to prove this formally/ where to begin.
I do know that $0 \leq |e^{-n|1-sin(x)|}|=|\frac{1}{e^{n|1-sin(x)|}}| \leq |1|$, but this doesn't show convergence in measure. Any help would be much appreciated, thanks.

$\endgroup$
  • $\begingroup$ The estimate at the very end of your question is wrong... $|\frac{1}{e^{n|1-\sin(x)|}}| \leq |\frac{1}{e^n}|$ does not hold true. $\endgroup$ – saz Nov 7 '18 at 7:16
  • $\begingroup$ @saz Oh I thought it was true because |1-sin(x)| is bounded between 0 and 2? $\endgroup$ – kemb Nov 7 '18 at 7:40
  • $\begingroup$ @saz Oh so I can only say that it is is $\leq 1$ my mistake. $\endgroup$ – kemb Nov 7 '18 at 7:40
1
$\begingroup$

Why don't you just observe that $f_n \to 0 $ except at a finite number of points (the points where $\sin \, x =1$) and almost everywhere convergence implies convergence in measure.

$\endgroup$
  • $\begingroup$ I see, makes sense. How would I prove formally that $f_n \rightarrow 0$ except the points where $|1-sin(x)|=0$. It is clear that the $f_n$ goes to 1 in this case ,but how would i show in all other cases it goes to 0 (seems obvious ,but not sure how to do it formally $\endgroup$ – kemb Nov 7 '18 at 7:44
  • $\begingroup$ $e^{-nt} \to 0$ if $t >0 $; $|1-\sin\, x|>0$ except at the points where $\sin\, x=1$. In the interval $[a,b]$ there sre only finite number of points where $|1-\sin\, x|=0$ (namely, numbers of the form $(2n+1)\pi /2)$. $\endgroup$ – Kabo Murphy Nov 7 '18 at 7:47
  • $\begingroup$ Sorry to bother you, But how do we know that the points where sin(x)=1 forms a set of measure 0 (in order to say that we have convergence almost everywhere). $\endgroup$ – kemb Nov 7 '18 at 19:15
  • $\begingroup$ @kemb $\sin\, x=0$ iff $x=n\pi$ for some integer $n$. Any countable set has Lebesgue measure $0$. $\endgroup$ – Kabo Murphy Nov 7 '18 at 23:18
  • $\begingroup$ I don't believe this actually works because convergence in measure only implies convergence in measure if the measure space is finite $\endgroup$ – kemb Dec 8 '18 at 8:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.