Proving two random variables differ with positive probability EDIT: Despite the help of the posters below, I'm still confused. I'm rephrasing the question slightly. Can someone hep me with rephrased problem:
Suppose that $X$ is a random vector and $Y$ a random variable that share some joint distribution. Suppose we know that the conditional density of $Y$ given $X$ is specified as
$$
f_Y(y|X)=\frac{1}{\sqrt{2\pi}\sigma_0}\exp\left(-\frac{1}{2\sigma_0^2}(y-X'\beta_0)^2\right).
$$
Here, $\sigma_0>0$ and $\beta_0$ is a vector of the same length as $X$. Taking log of the above expression, we have:
$$
\log f_Y(y|X)=-\frac{1}{2}\log(2\pi)-\frac{1}{2}\log(\sigma_0^2)-\frac{1}{2\sigma_0^2}(y-X'\beta_0)^2.
$$
Now, consider a function $g(y,x,\sigma,\beta)$ defined as
$$
g(y,x,\sigma,\beta)=-\frac{1}{2}\log(2\pi)-\frac{1}{2}\log(\sigma^2)-\frac{1}{2\sigma^2}(y-x'\beta)^2.\tag{$\star\star$}
$$
Note that $g(y,X,\sigma_0,\beta_0)$ agrees with $\log f_y(y|X)$. Prove that

$$ \sigma\neq\sigma_0\implies\Pr[g(Y,X,\sigma,\beta)\neq
 g(Y,X,\sigma_0,\beta_0)]>0. $$

p.s.
Original version:
Suppose that conditional on $x$, $y$ is normal with mean $x'\beta_0$ and variance $\sigma_0^2$. The log of the conditional density is then
$$
-\frac{1}{2}\log(2\pi)-\frac{1}{2}\log(\sigma_0^2)-\frac{1}{2\sigma_0^2}(y-x'\beta_0)^2.
$$
Now let $f(y\mid x;\beta,\sigma^2)$ denote the hypothetical conditional density of $y$ given $x$. That is, we may think of this as function of the random entities $y$ and $x$ satisfying
$$
\log f(y\mid x;\beta,\sigma^2)=-\frac{1}{2}\log(2\pi)-\frac{1}{2}\log(\sigma^2)-\frac{1}{2\sigma^2}(y-x'\beta)^2.\tag{$\star$}
$$
In particular, $f(y\mid x;\beta_0,\sigma_0^2)$ corresponds to the true conditional density of $y$ given $x$. Suppose $E(xx')$ is nonsingular. How can one show that if $\sigma\neq\sigma_0$, then
$$
\log f(y\mid x;\beta,\sigma^2)\neq \log f(y\mid x;\beta_0,\sigma_0^2)
$$
with positive probability? The book I am reading claims this in passing and I'd like to justify it but I can't proceed besides having a suspicion that we need to use ($\star$) somehow.
 A: If you have two continuous random variables with densities $f_1(x)$ and $f_2(x)$ then the probability they are the same is less than or equal to $$\int_{-\infty}^{\infty}\min(f_1(x),f_2(x))\,dx$$ which is less than $1$, unless $f_1(x)\not=f_2(x)$ only a set of measure zero.
If you take the difference between your two log densities, then you will have a function of $y$ which is $\frac{1}{2}\log(\sigma^2)-\frac{1}{2}\log(\sigma_0^2)+\frac{1}{2\sigma^2}(y-x'\beta)^2-\frac{1}{2\sigma_0^2}(y-x'\beta_0)^2$, i.e. a quadratic function of $y$ which is non-zero for all except at most two values of $y$ unless $\beta=\beta_0$ and $\sigma^2=\sigma_0^2$.  
A: Given $X=x$ you know that $Y$ is normally distributed with parameters mean $μ=x'β_0$ and variance $σ_0^2$, in symbols $$Y|X=x \sim N(x'β_0,σ_0^2)$$ Now for any $δ \in \mathbb R$ you have that $$F_{Y\mid Χ=x}(y\mid X=x)=P(Y\le δ\mid X=x)=P\left(Z\le\frac{δ-x'β_0}{σ_0}\right)=Φ\left(\frac{δ-x'β_0}{σ_0}\right)$$ If $σ\neq σ_0$, and using that $Φ$ is a monotone increasing function (and therefore injective) you obtain (unless $δ-x'β_0=0$) that $$\Phi\left(\frac{δ-x'β_0}{\color{blue}{σ_0}}\right)\neq\Phi\left(\frac{δ-x'β_0}{\color{blue}σ}\right)$$ or equivalently that $$\int_{-\infty}^{δ}f_{Y\mid X=x}(y\mid X,\color{blue}{σ_0},β_0)dy\neq\int_{-\infty}^{δ}f_{Y\mid X=x}(y\mid X,\color{blue}σ,β_0)dy$$ for every $δ \in \mathbb R$. 

Actually the parameter $σ^2$ is the variance of the conditional distribution of $Y$ given $X=x$. Changing this parameter changes the shape of the normal curve and therefore it is immediate that these two functions differ completely. (This point is actually mentioned in another answer). 
A: To Henry: I deleted my comment because I was (am) still a bit confused. Could you (or anyone) help me verify this argument? I would mark Henry's post as the answer once I understand this more.
Let $\theta=(\beta,\sigma^2)$, $\theta_0=(\beta_0,\sigma_0^2)$, and $a(y,x;\theta)=I[\log f(y\mid x;\theta)=\log f(y\mid x;\theta_0)]$ where $I$ denotes the indicator function. Then,
$$
E[a(y,x;\theta)\mid x]=\Pr\Big[\log f(y\mid x;\theta)=\log f(y\mid x;\theta_0)\Big|x\Big]=0
$$
because as in Henry's post, given $x$ and $\sigma\neq\sigma_0$, $\log f(y\mid x;\theta)=\log f(y\mid x;\theta_0)$ for at most two points of $y$ so the event of that carries $0$ probability (recall that given $x$, $y$ follows a normal distribution). It follows that
$$
\Pr\Big[\log f(y\mid x;\theta)=\log f(y\mid x;\theta_0)\Big]=E[a(y,x;\theta)]=E\Big[E[a(y,x;\theta)\mid x]\Big]=E[0]=0.
$$
A: Suppose the measure of the set of values of $y$ at which which the two densities differ is zero.  Then the two distributions assign the same probability to every Borel set.  That means they're the same distribution.  The entails that the have the same expected value and the same variance.
