Let $(X,M.\mu)$ be a measure space and let $f\colon X \rightarrow \mathbb{R}$ be a measurable function. Show that $$\lim_{n \to \infty} \int_X \left(1-\left(\frac{2}{e^{f(x)}+e^{-f(x)}}\right)^n \right)d\mu=\mu(\{x \in X:f(x)\neq0\})$$

My attempt:

I think its an application of Dominated or Monotone Convergence Theorem followed by Chebyshev's inequality. I didn't able to dominate the function inside the integral with an integrable function(so, i guess its not DCT) and even if I try to use MCT, I can't use it. The thing inside the integral is $\left(1-\frac{1}{\cosh(f(x))}\right)^n$ and I only know that $\cosh$ is increasing from $[0,\infty)$ but its decreasing from $(-\infty,0]$.

Any help is appreciated!

  • $\begingroup$ the integrand is bounded above by 1, why cannot you use DCT? $\endgroup$ – ablmf Dec 24 '18 at 20:13
  • $\begingroup$ @ablmf Well, is the constant $1$ function integrable? That may not fly when $\mu(X)=\infty$. $\endgroup$ – Clement C. Dec 24 '18 at 20:43
  • $\begingroup$ I am, however, very confused as to why and where Chebyshev's inequality would come into play. $\endgroup$ – Clement C. Dec 24 '18 at 20:48

Hint: Let $f_n$ be the $n$th integrand. Then $0\le f_1\le f_2 \le \cdots $


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