Uniform distribution over the real using a bijection from [0,1]?

Question

Probably a dumb question but I was struggling to reconcile the fact that a uniform distribution over the reals is impossible with the fact that we can define a uniform over a finite subset of the reals and then put this subset in bijection with the entire set of reals.

For example if we have P(x) = U(0,1), can we not simply define Q(x) as, say, P(tan^-1(x)), which will just "stretch out" the distribution P. It seems like for every point in [0, $ \mathbb R^+ $), it will be mapped onto a point in [0,1] and will therefore by definition have density of 1, so we end up with a uniform over the reals with density 1 everywhere.

I understand that this is impossible and I also see how intuitively it doesn't look like a real because intervals of equal length in [0,1] don't get mapped onto equal intervals in [0, $ \mathbb R^+ $). But I'm not seeing the formal reasoning behind this since surely all finite intervals have measure 0 anyway so they are equal

My gut feel is that the answer is something like "There's no guarantee that the property of beng a valid distribution of preserved under bijection etc. So go ahead and define your Q(x), it wouldn't automatically make it a distribution". Is this on the right track?

Thanks!

Answer 1

What you describe does not define a uniform distribution, since the measure of an interval $[x,x+1]$ far from 0 will be very small, as its preimage in $[0,1]$ is $[tan^{-1} x, tan^{-1} (x+1)]$. For a uniform distribution you would need that the derivative of your function be constant.