Can a density function in a closed ball have an unbounded expected value? Given a closed ball, $$\mathcal{F}=\{g:D(g,f)\leq\epsilon\}$$ for a distance measure, $$D(g,f)=\int_{-\infty}^\infty g(y)\log\frac{g(y)}{f(y)}\mathrm{d}y$$ is there any density function $g$, which has an expected value equals $\infty$? 
Notes: $f:\mathbb{R}\rightarrow\mathbb{R}$ and $g:\mathbb{R}\rightarrow\mathbb{R}$ are density functions with $g(y)>0$ and $f(y)>0$. The expected value w.r.t. $f$ is known to be $<\infty$. The density function $f$ is given and known.
Thank you very much..
 A: It seems difficult to expect that $g$ being close to $f$ for the entropy and $f$ being integrable could imply an integrability property of $g$. Assume for example that $f$ and $g$ are density probabilities on $(1,+\infty)$, with $$f(x)=\frac2{x^3},\qquad g(x)=(1-a)\frac2{x^3}+a\frac1{x^2},
$$
for every $x\gt1$, for some $a$ in $(0,1)$. Then $f$ is the density of an integrable distribution, $g$ is the density of a non integrable distribution, and
$$
D(g,f)=\int_1^\infty\left((1-a)\frac2{x^3}+a\frac1{x^2}\right)\,\log\left(1-a+\frac12ax\right)\,\mathrm dx.
$$
An integration by parts based on the functions $u$ and $v$ defined on $(1,+\infty)$ by
$$
u(x)=(1-a)\frac1{x^2}+a\frac1{x},\qquad v(x)=\log\left(1-a+\frac12ax\right),
$$
yields
$$
D(f,g)=\left.-uv\right|_1^\infty+\int_1^\infty uv'=\log\left(1-\frac12a\right)+a\int_1^\infty\frac{1-a+ax}{2(1-a)+ax}\,\frac{\mathrm dx}{x^2}
$$
The fraction in the last integral is at most $1$ hence
$$
D(f,g)\leqslant\log\left(1-\frac12a\right)+a\leqslant\frac12a.
$$
In particular, $D(f,g)\to0$ when $a\to0$ although the distribution with density $g$ is not integrable, for any positive $a$.
