prove that $\int _{-\infty }^\infty \frac{1}{1+f(x)}\operatorname{dx}$ diverges

Question

Let $f : \Bbb R \to [0,\infty )$ be a measurable function. If $\int _{-\infty }^\infty f(x)\operatorname{d}x= 1 $ then prove that $\int _{-\infty }^\infty \frac{1}{1+f(x)}\operatorname{d}x= \infty $.

I am absolutely clueless on how to start this except applying the definition of a measurable function.

Any hints on this problem will be highly helpful.

Answer 1

Here's one way to do it:

Let $I_n = \frac{1}{2n} \int_{-n}^n f(x)\,dx$. By monotone or dominated convergence, $I_n \to 0$.

By Jensen's inequality, $$\frac{1}{2n} \int_{-n}^n \frac{1}{1+f(x)}\,dx \ge \frac{1}{1 + I_n}.$$ That is, $$\int_{-n}^n \frac{1}{1+f(x)}\,dx \ge \frac{2n}{1+I_n}.$$ Now let $n \to \infty$.

Answer 2

Since $f(x)\ge0$, we have $\frac{f(x)}{1+f(x)}\le f(x)$. Therefore, $$ \begin{align} \int_{-n}^n\frac{\mathrm{d}x}{1+f(x)} &=2n-\int_{-n}^n\frac{f(x)\,\mathrm{d}x}{1+f(x)}\\[6pt] &\ge2n-1 \end{align} $$