Variable transformation under a discontinuous mapping function introduce different bias? For any real number $x \in R$, consider its randomized version $y \sim \mathcal{N}(x,1)$. Now consider a mapping function of $y$ denote by $f(y)$.  We want to study the following bias term $R(x) = | E[f(y)] - f(x)  |$. My hypothesis here is that if $f$ is a discontinuous function, then $R(x)$ is not a constant w.r.t $x$. I do not know if this hypothesis is correct, or any condition on $f$ to make my hypothesis correct. If it is correct, can you shed some light how to prove this.
For example, if I choose $f = \mathbb{1}\{y \geq 0\}$ (the threshold function), which returns 1 if $y \geq 0$ and return $0$ if $y <0$,   $f$ is a discontinuous  function at $y =0$. Then for x = -10000 or x = 10000 $R(x)$ should be very small, because $-10000, 10000$ is very far from the discontinuous point $y=0$. However, for $x=0.001$ $R(x)$ should be larger because $x=0.001$ is very close to the discontinuous point.
Thanks
 A: Your hypothesis is correct.
So that we may be notationally clear, let $x$ be a number and let $Y_x$ be its randomized version.  In terms of the density function for a standard Normal variable, $\phi,$ this means $Y_x$ has the density function $y \to \phi(y-x) = \phi(x-y).$  Given any (measurable) mapping $f:\mathbb{R}\to\mathbb{R},$ in terms of the convolution operator $\star$ we may write
$$E[f(Y_x) - f(x)] = E[f(Y_x)] - f(x) = \int_{\mathbb{R}} f(y)\phi(x-y)\,\mathrm{d}y - f(x) = (f\star \phi)(x) - f(x).$$
$R(x)$ is defined to be the absolute value of this expression.  Suppose $R$ is a constant function $R(x)=r$ for all $x.$  Then
$$r = R(x) = \pm\left((f\star \phi)(x) - f(x)\right)$$
implies
$$f(x) \pm r = (f\star \phi)(x).$$
The function $f\star \phi$ is at least as differentiable as $\phi$ is (because we may differentiate it by differentiating $\phi$ in the defining integral).  In particular, because $\phi$ is differentiable, $f\star \phi$ is differentiable, whence it is continuous. Since $f \pm r$ is continuous and the constant function $r$ is continuous, so is $f.$
The contrapositive of this implication is

When $f$ is not continuous, $R$ cannot be constant.

