In variational autoencoders, what does p(x|z) mean? If $x \sim \mathcal{N}(\mu,\,\sigma^{2})$, then it is a continuous variable, and therefore $P(x) = 0$ for any x. One can only consider things like $P(x&lt;X)$ to get a probability greater than 0.
So what is the meaning of probabilities such as $P(x|z)$ in variational autoencoders? I can't think of $P(x|z)$ as meaning $P(x&lt;X|z)$, if $x$ is an image, since $x&lt;X$ don't really make sense (all images smaller than a given one?)
Whilst you're right that for any continuous distribution $P(X = x) = 0 \;; \forall x \in \mathcal{X}$ where $\mathcal{X}$ is there support of the distribution, they are not referring to probabilities here, rather they are referring to density functions (though this should really be denoted with a lower case $p$ to avoid confusion such as this).
$p(x|z)$ is a conditional distribution, which is also allowed in the continuous case -- you can also 'mix and match', i.e. $x$ could be continuous and $z$ could be discrete, and vice-versa.
In the paper, all the authors are meaning when they write $p(x|z)$ is the density of $x$ conditioned on $z$; in VAE's with an image application this is the conditional density of the image $x$ given your latent vector $z$.