# An Application of Convergence theorem

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!

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

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