Even if your probability measure is absolutely continuous with respect to Lebesgue measure on $Omega={mathbb R}^2$, I don't think this suffices for the function $f$ you have defined to be continuous (just take $X$ to be independent from $Y$, i.e. your probability measure is just the product of two probability measures, "one on each axis", and choose the one for $X$ to be something in $L^1({mathbb R}, {mathcal B}, dx)$ which is discontinuous.
On the other hand, if ${mathbb P}$ is not just absolutely continuous with respect to Lebesgue measure on $Omega$, but has a continuous density function wrt said measure, then your function $f$ will be continuous -- just because integrating over a ball of radius $eta$ with centre $eta$ can only smooth things out, so that continuity of the original density function goes over to continuity of your conditional probability. [This is a fairly straightforward observation using basic properties of usual integration in the plane.]
So in your example, the Gaussian structure isn't really relevant as far as I can see. Also, if the original density function is strictly positive on the cylinder ${(x,y) : |y-y_0|<eta }$, then the conditional probability you've defined will also be strictly positive; this condition is evidently not necessary, but I suspect in the examples you're interested in something like it should hold.
In between these two extremes, I'm not sure what else one can say. Perhaps, from your point of view, it's more important to go up to infinite-dimensional $Omega$ but place restrictions on the kind of probability measure which you wish to consider.
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