In the real world, the ultraviolet catastrophe doesn't happen because the quantization of photons modifies the classical behavior of light at frequencies comparable to and higher than the temperature. But classical electromagnetism is a mathematically self-consistent theory, so we could imagine a world where $\hbar = 0$ and electromagnetism remains classical to arbitrarily high frequencies. How would the ultraviolet catastrophe work in such a world? Classically, all frequencies are equally populated at all temperatures and each has energy $\frac{1}{2} k_B T$, apparently leading to infinite energy radiation at any temperature, which doesn't seem compatible with conservation of energy. What would happen if you put a theoretical fully classical system in contact with a thermal bath (which seems like a physically reasonable set-up)?
My guess is that since this system has an infinite number of quadratic degrees of freedom, the usual canonical-ensemble derivation of thermal equilibrium breaks down. I think that if you couple a perfectly conducing cavity to any thermal bath, no matter how large, then the cavity will absorb an unboundedly large amount of energy from the bath. In the usual derivation, we assume that the bath has so much more energy than the system that its energy density is independent of the state of the system, but in this case that assumption will eventually be violated. The cavity's energy will become comparable to the bath's, so to find its equilibrium state, we will need to consider the details of the bath and treat the combined cavity-bath system in the microcanonical ensemble. So the Boltzmann distribution and the equipartition theorem will no longer apply, and the ultraviolet catastrophe will be avoided.
Answer
I think this is an interesting question. If one tries to couple the EM field to matter somewhat realistically and classically, he needs a model for matter. One possible model would be that of an assembly of uncorrelated dipoles. If we assume these dipoles to be point dipoles then the energy they emit is related to the wavelength as $\varepsilon_{\lambda} \sim \lambda^{-4}$. There is no discussion that this is correct as this corresponds to the Rayleigh scattering result and we more or less witness its effect everyday when we go outside.
So, it could be that the "ultraviolet catastrophe" attributed to the Rayleigh-Jeans formula, which incidentally has also an EM energy density that goes as $\sim \lambda^{-4}$, assumes somewhere that the scattering material objects are point-like.
At the very least for consistency reason, at equilibrium, the energy density emitted by an assembly of dipoles has to agree with that calculated independently for the EM field and the two above results have to agree.
Now, the point is that at high frequency, the wavelength becomes of the order of the size of the scatterers and we could get for instance something close to Mie scattering instead of Rayleigh scattering. The effect it would have is that the energy fraction emitted by a scatterer becomes roughly independent of the wavelength in this regime and so something will probably happen and prevent a fully diverging energy density at short wavelength.
So in effect, one would have to cut off the spectrum somewhere and make it at least saturate at some value.
Note that this is something ubiquitous in statistical mechanics (on a quite different note the partition function of a single hydrogen atom diverges and one needs often to put by hand a cutoff corresponding to an atom radius big enough to create an ambiguity between a pair of H atoms and an H2 molecule) and it is probable that people are often too eager to claim, a posteriori, that a new theory was definitely needed while something could possibly have been done with a better model.
This of course does not mean that quantum mechanics is not required in the end but it means that the dramatic failure of the Rayleigh-Jeans formula is a problem owing to all the assumptions leading to the said formula, not just one.
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