Let's say I have simple BPSK scheme and a receiver RF stage such that I have an ideal analogue bandpass filter with bandwidth \$ f_{bw} \$ and then a down converter and finally a sampler. I am trying to use oversampling so that I can integrate the inputs through a correlator and increase the overall SNR of the received bits.
I know for sure if the input to the correlator has AWGN noise this will work. However, my friend claims the output of the sampler will have correlated noise and oversampling will not work, due to ideal analogue filter which will essentially make the AWGN noise coloured at the analogue domain. He tries to prove this through the Fourier transform of the noise power at the output of the analogue filter.
I simply cannot grasp this idea, sure the noise at the output of the analogue filter will be also bandlimited but we aren't sampling the Fourier transform of the noise, we are sampling the time domain version of it, once I sample it, the noise power should be distributed all over the frequency axis once it is sampled with oversampling. My noise power should stay the same but the noise should be redistributed to the whole frequency axis of concern after sampling.
His other claim is that my type of thinking breaks Shannon's capacity theorem since I can just oversample infinitely and integrate them to achieve infinite SNR (at least for BPSK). But shouldn't increase in the sampling rate basically means using more bandwidth hence it is consistent with Shannon's channel capacity even though transmitter sampling rate was low but receiver oversamples and then integrates the samples.
I am confident with my thought process but I am not able to find a proper source to prove or disprove them.