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?
 A: Suppose that $f$ is nonnegative (you say it's a distribution?). Notice that $1=\frac{1}{1+f}+\frac{f}{1+f}$. Also $0\leq \frac{f}{1+f}\leq f$ which means $\frac{f}{1+f}$ is integrable. Since the l.h.s. is not integrable, it follows that $\frac{1}{1+f}$ is not integrable. 
A: 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
A: An obvious thing to do with the hint is observe that $f > 1$ is the same thing as $\frac{1}{1+f} < \frac{1}{2}$.
A: Here I assume two things: $\int |f|<\infty$ and $f>-1$ a.e.
\begin{aligned} \int_{-\infty}^\infty \frac{1}{1+f} dx & \geq \int_{-\infty}^\infty \frac{1}{1+|f|} dx \\ & = \int_{-\infty}^\infty 1-1+\frac{1}{1+|f|} dx \\
 & = \int_{-\infty}^\infty 1-\frac{|f|}{1+|f|} dx \\
& \geq \int_{-\infty}^\infty 1-|f| dx \\
& = \int_{-\infty}^\infty 1 - \int_{-\infty}^\infty |f| \\
& = \infty.
\end{aligned}
A: I assume that:
$$\int_{-N}^{N} \frac{1}{f(x)+1} dx < +\infty,$$
for any $N > 0$ since $f(x) > 0 ~\forall x$.
Furthermore, I assume $$\lim_{x \to \pm\infty} f(x) = 0$$.
This means that:
$$\forall \varepsilon > 0 ~\exists N > 0: |x| >N \Rightarrow f(x) < \varepsilon.$$
Then:
$$f(x) < \varepsilon \Rightarrow \\
f(x) + 1 <  \varepsilon + 1 \Rightarrow \\
\frac{1}{f(x)+1} > \frac{1}{\varepsilon + 1} ~\forall x:|x| > N.
$$
This implies that:
$$\int_{N}^{+\infty} \frac{1}{f(x)+1} dx > \int_{N}^{+\infty} \frac{1}{\varepsilon+1} dx = +\infty,$$
and
$$\int_{-\infty}^{-N} \frac{1}{f(x)+1} dx > \int_{-\infty}^{N} \frac{1}{\varepsilon+1} dx = +\infty.$$
Since:
$$\int_{-\infty}^{+\infty} \frac{1}{f(x)+1} dx = \int_{-\infty}^{-N} \frac{1}{f(x)+1}dx + \int_{-N}^{N} \frac{1}{f(x)+1} dx + \int_{N}^{+\infty} \frac{1}{f(x)+1} dx$$
we can conclude that $\int_{-\infty}^{+\infty} \frac{1}{f(x)+1} dx $ diverges.
