Of I'll say something about how to do this but read on afterwards--I think it might not be a good idea in general. The best way to do it might be to decide first what level of precision you're after, and then calculate over the interval that would round to that value. So if you really want to know the probability of observing a value of .01, given that any "true" value in [.0095, .0105) would round to .01, then calculate that integral. But if you're rounding more precisely, you might want to look at [.00995, .01005) or whatever. Of course this is a smaller interval, and the probability will be smaller.

This sort of thing is done when using the normal distribution, which is continuous, to approximate the binomial distribution, which is discrete and takes only positive integer values. We might approximate the chance that the variable takes value 500 by integrating the normal pdf from 495.5 to 500.5.

The problem with doing this to find probabilities of pseudo-discrete RVs, like rounded values of continuous RVs, is that the more precise you get, the smaller the interval is and the smaller the probability gets, eventually shrinking to 0. Suppose you have a standard normal RV. The pdf is highest at the mean, 0. But if you calculate the probability of X=0 by integrating the pdf from .99995 to 0.00005, it'll be really tiny. You need to be careful in this case to realize that it's still the most likely value, even though all values have probability 0, and the estimates all go to 0 as well. This is a little confusing, but something you need to accept in probability very early on is that intuition is not to be trusted. There are very specific definitions of probability, and it's perfectly acceptable to say that the probability of a continuous RV taking value x is 0 for every x, even though it has to take *some* value.