"an irrational angle" like $2\pi$?

@md2perpe thanks for pointing it, clarified. Irrational degrees, of course. Not radians.

This “paradox” doesn’t mean anything at all. Your set has measure zero to start with. In general, if you start with a set of finite measure $A$ and rotate it to $B$, where $B$ is a strict subset of $A$, then it just tells you $A \setminus B$ is null. Nothing more.

@DavidGao I did not mention Lebesgue measure at all. Neither I meant it. Why are you talking about it?

@Anixx Because the Lebesgue measure is the unique (up to scaling) measure that is invariant under rotation, reflection, and translation? That’s the entire reason why Banach-Tarski paradox is surprising in the first place. And in any case the second half of my comment applies equally well to any finitely-additive rotation-invariant measure.

Also, the Wikipedia article on Banach-Tarski paradox has a section specifically mentioning why there is no Banach-Tarski paradox on a plane - there is, in fact, a finitely-additive translation- and rotation-invariant measure defined for all subsets of $\mathbb{R}^2$ extending the Lebesgue measure, so no paradoxical decomposition as in the Banach-Tarski paradox can exist in the plane. (And, again, this doesn’t contradict your “paradox” because you sets have measure zero.)

@DavidGao if B is a strict subset of A, then A \ B cannot be null, no? As to your other comment, my point is, this paradox is already surprising in 2D case and without Axiom of Choice, if we want to have a finitely-additive (for bounded sets) and rotation-invariant measure. The unbounded case is already paradoxical in 1D case (Hilbert's hotel paradox).

@Anixx There are plenty of nonempty null sets, what do you mean? (Null means measure zero, not empty, as is standard terminology in measure theory.)

@DavidGao the point is it is null in your terminology only under Lebesgue measure. If we want to chose another measure, it would not necessarily be null. Particularly, if we want our measure to be non-null for any non-empty set.

@Anixx I don’t find it surprising at all, but maybe your perspective is different. Though, again, it doesn’t put any restriction on finitely-additive rotation-invariant measures on the plane, as mentioned in my previous comments. It just tells you the $A \setminus B$ (the singleton set of your initial point in your case) has measure zero w.r.t. this finitely-additive rotation-invariant measure, is all. The Banach-Tarski paradox is only surprising because you get something of a different volume at the end, which your example does not do.

@DavidGao suppose, we want our measure to be non-null for any non-empty set. Then we get this paradox and need do something, like introduce non-measurable sets or make it non-rotation-invariant.

@Anixx No, just take the counting measure. You need something like bounded sets have finite measures, but then that’s completely obvious in the first place. If you have a point with positive measure, you can just rotate it to get infinitely many points with the same positive measure in a bounded set, which cannot happen if you require bounded sets to have finite measures. There’s really nothing interesting or surprising about this, and I don’t think it makes sense to ask non-empty sets to always have positive measure. That’s simply counterintuitive.

@DavidGao why bounded sets should have finite measures?...

@DavidGao my point is, in 1D case due to Hilbert's hotel paradox we have to make our measure non-translation-invariant, and this is OK. In 2D case, we can fix the things by making the measure non-rotation invariant, but it seems too much. And in 3D case, only a theory of surreal integration will help...

(You can do the exact same thing and see that any finitely-additive translation-invariant measure on the real line that evaluates to finite numbers on bounded sets must have all singletons being of measure zero. Which, I’d say, anyone working with these concepts should find quite intuitive.)

@DavidGao but why should it evaluate to finite numbers? I prefer surreal numbers.

@Anixx If bounded sets don’t need to have finite measures, then what is your point of referring to bounded sets at all? It’s even less of a “paradox” that way. Just take the counting measure. It is invariant under literally any bijection.

@DavidGao Hilbert's hotel paradox happens only with unbounded sets: a set can be made into a strict subset of itself by translation. With bounded sets this is impossible in 1D case.

@Anixx … do you mean hyperreal? Surreal numbers form a proper class. There’re not that many subsets of reals (or the real plane, or the $3$-dimensional space) for you to have any reason to use surreal numbers.

@Anixx I was not referring to Hilbert’s hotel paradox. By “exact same thing” I meant the (far simpler) argument I gave in my comments showing the exact same thing, namely, singletons have to be null for your measure to satisfy certain conditions.

@DavidGao I mean surreals belonging to $No(\omega_2)$, which includes countable surreals and continuum. Those are enough for an Euclidean space.

@DavidGao in 1D case there is no paradox besides the Hilbert's hotel, which can be properly mitigated by making the measure non-translation-invariant regarding non-bounded sets.

@Anixx Hilbert’s hotel paradox is hardly a “paradox” in the first place, nor, again, was I using it in any form. I have no idea what you are talking about.

@Anixx And, you might as well just use cardinal numbers and counting measure given your specification. I don’t see what exactly is the meaning of using surreal for this. If you have a specific idea of using surreals in mind, actually write it down and ask a question (or write a paper/blog post/whatever) for that. You’re hardly making any sense right now.

@DavidGao Hilbert's hotel prohibits a translation-invariant finitely-additive measure that is non-zero for any non-zero set. That's why we have to make it non-translation invariant for nonbounded sets.

@Anixx No it doesn’t, the counting measure is finitely-additive (in fact, arbitrarily additive), translation-invariant, and non-zero for any non-empty set.

@DavidGao a measure based on cardinal numbers would not be additive.

@Anixx … what? Counting measure is based on cardinal numbers and perfectly additive. What are you talking about…?

@DavidGao we want our measure also be ordered in this way: $\mu(A\cup B)>\mu(A)$ iff $B$ is not an empty set.

..and $A∩B$ is not a zero set.

@Anixx Then you’re out of luck, I guess. Even surreal numbers (or any other ordered field, for that matter) can’t help you, as it is immediately obvious that if the measure is finitely-additive and translation-invariant (on the real line/plane/space, I don’t care), then your condition cannot be satisfied.

In any case, I think this discussion has gone on long enough, and you’re still making so little sense. I don’t think any further response from me is useful, so I won’t respond any further. If you have a specific idea as to how to use surreal numbers, ask a question about that, instead of hiding it in a question that ostensibly has nothing to do with surreal numbers.

@DavidGao There are no contradictions in the 1D case besides Hilbert's hotel, but as I said it can be mitigated by dropping translation-invariance regarding unbounded sets.