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.
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.