properties of distributions If $$\int_{-\infty}^\infty f dx = 1$$, with $f > 0 \forall x$, then prove or disprove: $$\int_{-\infty}^\infty \frac{1}{1 + f} dx $$ diverges.
The hint I got is to consider the measure of the set$(x:f > 1)$. May be the measure is zero thereby ensuring the divergence of integral?

If $$ \int _{-\infty }^{\infty}\frac{1}{f+1} d x < \infty$$
Then $$\int _{-\infty }^{\infty}\frac{1/4}{f+1} d x +\int _{-\infty }^{\infty}f d x < \infty$$
But
\begin{align*}
\int _{-\infty }^{\infty}\frac{1/4}{f+1} d x +\int _{-\infty }^{\infty}f d x&=
\int _{-\infty }^{\infty}\frac{1/4}{f+1}+f d x\\
&=\int _{-\infty }^{\infty}\frac{f^2+ f+1/4 }{f+1} d x \\
&=\int _{-\infty }^{\infty}\frac{(f+1/2)^2}{f+1} d x \\
&\geq \int _{-\infty }^{\infty}\frac{(f/2+1/2)^2}{f+1} d x\,\,\text{ since } f\geq 0\\
&\geq 1/4\int _{-\infty }^{\infty}\frac{(f+1)^2}{f+1}f d x \\
&\geq 1/4\int _{-\infty }^{\infty}f+1d x \\
&>\infty
\end{align*}
A contradiction
