Q:

If $x\in\mathbb R^n$ and $f:\mathbb R→(0,∞)$, how can we show that $\sum_{i=1}^n(\ln f(y_i)-\ln f(x_i))≠0$ for Lebesgue almost all $y\in\mathbb R^n$?

Let

$f:\mathbb R\to(0,\infty)$ with $f>0$ and $$\int f(x)\:{\rm d}x=1$$ and $g:=\ln f$
$n\in\mathbb N$ and $$s(x,y):=\sum_{i=1}^n\left(g(y_i)-g(x_i)\right)\;\;\;\text{for }x,y\in\mathbb R^n$$
$x\in\mathbb R^n$ and $Y$ be a $\mathbb R^n$-valued normally distributed random variable on a probability space $(\Omega,\mathcal A,\operatorname P)$ with mean vector $x$ and covariance matrix $\sigma I_n$ for some $\sigma>0$

I want to show that $$\operatorname P\left[s(x,Y)=0\right]=0.$$

As the $n$-dimensional normal distribution has a (strictly) positive density with respect to the Lebesgue measure $\lambda^n$ on $\mathbb R^n$, it should be sufficient to show that $$s(x,y)\ne0\;\;\;\text{for }\lambda^n\text{-almost all }y\in\mathbb R^n\tag1.$$ How can we do that?

A:

Let $n=1$ and $f(x)=\frac  1 2 $ for $|x| \leq 1,\frac 1 {4x^{2}}$ for $|x| >1$. Then $\lambda \{y: f (y)=f(0)\}=2>0$. So the claim is  false.

