Conditional Distribution of Normally distributed random variable Let $x \sim \mathcal{N}\left(0,\sigma^2\right)$ and $y = x+\epsilon$ where $\epsilon\sim \mathcal{N}\left(0,\sigma^2_\epsilon\right)$ and independent of $x$. We know that the conditional distribution of $x$ given $y$ is $$x\mid y \sim \mathcal{N}\left( \frac{\sigma^{2}}{\sigma^{2}+\sigma_\epsilon^{2}}y , \frac{\sigma^{2}\sigma_\epsilon^{2}}{\sigma^{2}+\sigma_\epsilon^{2}}\right)$$
My question is that: How can we compute the posterior distribution of $x$ given $y\leq\overline{y}$? That is, the exact value of $y$ is unknown, we only know that the event $y \in \left[ -\infty, \overline{y} \right]$.
My attempt was to use the Bayes' rule: $$f\left(x|y\leq \overline{y}\right)=\frac{f\left(y|y\leq \overline{y},x\right)f\left(x\right)}{f\left(y|y\leq \overline{y}\right)}$$
but I remain unsure if this is the right approach. Thanks in advance for the help!

If the observation is that$$Z=\mathbb I_{Y\le\bar y}$$instead of $Y=y$ itself, the conditional distribution of $X$ is derived from the joint distribution of $(X,Z)$:
$$f(x|z)=\dfrac{f(x,z)}{f(z)}=\dfrac{\mathbb P(Z=z|X=x)f(x)}{\mathbb P(Z=z}$$
with$$\mathbb P(Z=z)=\mathbb P(Y\le\bar y)=\Phi(\bar y/\sqrt{\sigma^2+\sigma^2_\epsilon})$$
and
$$\mathbb P(Z=z|X=\mathbb P(Y<\bar y|X=x)x)=\Phi(\bar y-x/\sigma_\epsilon)$$
