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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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