Integral of a measurable function

Question

I do not know what should i keep as title for this question...

Question goes like this..

Let $f:\mathbb{R}\rightarrow [0,\infty)$ be a measurable function. If $\int_{-\infty}^{\infty}f(x)dx=1$ prove that

$$\int_{-\infty}^{\infty}\frac{1}{1+f(x)}dx=\infty.$$

Hint that was given is : First show that $\{x:f(x)<1\}$ is of infinte measure.

I do not know how to proceed.. All that i could think about this is the following.

By $\int_{-\infty}^{\infty}f(x)dx=1$ i see that $\lim_{n\rightarrow \infty}\int_{-n}^n f(x)dx=1$ i.e., given $\epsilon>0$ there exists $N\in \mathbb{N}$ such that for all $n\geq N$ we have

$$\left|\int_{-n}^nf(x)dx-1\right|<\epsilon$$

I do not see any thing more than this..

Answer 1

For the hint, suppose that $\{x:f(x)<1\}$ had finite measure. Then $\{x:f(x)\geq 1\}$ has infinite measure. Then, $\{x:f(x)\geq 1\}$ has a subset $A$ of measure $2$. But then, $$ \int_{\mathbb{R}}f(x)dx\geq\int_Af(x)dx\geq\int_Adx=\mu(A)=2>1. $$

Since $\{x:f(x)<1\}$ has infinite measure, use the hint of @G.Sassatelli (a very similar argument to the argument above) to finish the proof.

Answer 2

For the hint, notice $$ 1=\int_{-\infty}^\infty f=\int_{f<1}f+\int_{f\geq1}f $$ Since $f\geq 0$, this implies $\int_{f< 1}f\geq 0$, while $\int_{f\geq1}f\geq\int_{f\geq1}1=m(\{x:f(x)\geq1\})$. Hence

$$ 1=\int_{-\infty}^\infty f=\int_{f<1}f+\int_{f\geq1}f>m(\{x:f(x)\geq1\}) $$

Using this and the hint given in the comment, you can conclude the proof.