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I would just like to ask whether my proof is correct. The problem is to show $f_n(x)=e^{-n|1-\sin x|}$ converges to $f(x)=0$ in measure on $[a,b]\in \mathbb R$

Here is my attempt, using Chebyshev's inequality and Dominated Convergence.

Let $\epsilon>0$ be arbitrary but fixed and $A_n=\{x \in [a,b]: e^{-n|1-\sin x|}> \epsilon\}$. We need to show $m(A_n)\to 0$. Now $m(A_n)\leq \frac{1}{\epsilon} \int_{[a,b]}e^{-n|1-\sin x|}dx$ by Chebyshev. But it is clear that $|e^{-n|1-\sin x|}|<1$ on $[a,b]$ for every n. Now 1 is integrable on [a,b], hence by dominated convergence, since for every fixed x, $e^{-n|1-\sin x|} \to 0$, then we can interchange limit and integral to get that $\int_{[a,b]}e^{-n|1-\sin x|}dx\to 0$ But $m(A_n)\leq \int_{[a,b]}e^{-n|1-\sin x|}dx$ for every n and so taking the limit we get $m(A_n)\leq 0$ which implies $m(A_n)=0$.

Can somebody tell me if my reasoning is correct? I feel I have a made a mistake somewhere. Thanks.

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    What about when $x$ is a multiple of $\pi$? – 2017-02-20
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    I don't see what would change? We would have $e^{-n}$ which still goes to 0. – 2017-02-20
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    Well, it is true that it converges in measure, I presumed you were asking about your proof? If $x$ is a multiple of $\pi$ it is not true that $e^{-n|1-sinx|} \to 0$ or $|e^{-n|1-sinx|}|<1$. These are technicalities, but necessary for a correct proof. – 2017-02-20
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    I think you mean when x is an odd multiple of pi/2? In that case we would have the measure of An less or equal to b-a but that's countable number of x's hence measure 0. Actually when x is $\pi/2 + 2\pi k$ for k a natural number ? – 2017-02-20
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    Yes, I meant odd multiples of ${\pi \over 2}$. I know it is true, you asked about your reasoning, I was just pointing out incomplete or incorrect points in your proof. – 2017-02-20
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    Thanks, so since that set also has measure 0, we can split any set $A_n$ into the set of odd multiples of pi/2 union with the rest of [a,b] and get the result? – 2017-02-20
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    Well, I would replace the $|e^{-n|1-\sin x|}|<1$ by $|e^{-n|1-\sin x|}| \le 1$ and to use the DCT, you only need $f_n(x) \to f(x)$ ae., so you only need to note that $e^{-n|1-\sin x|} \to 0$ ae. – 2017-02-20
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    Oh yes, I did not notice that I only had strict inequality. Thank you. – 2017-02-20

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