Prove Sequence of Functions Converges in Measure

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 $$n \rightarrow \infty$$. exactly 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.

• 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. – saz Nov 7 '18 at 7:16
• @saz Oh I thought it was true because |1-sin(x)| is bounded between 0 and 2? – kemb Nov 7 '18 at 7:40
• @saz Oh so I can only say that it is is $\leq 1$ my mistake. – 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.
• 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 – kemb Nov 7 '18 at 7:44
• $e^{-nt} \to 0$ 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 $(2n+1)\pi /2)$. – Kavi Rama Murthy Nov 7 '18 at 7:47
• @kemb $\sin\, x=0$ iff $x=n\pi$ for some integer $n$. Any countable set has Lebesgue measure $0$. – Kavi Rama Murthy Nov 7 '18 at 23:18