G'day Alex and gary.
Don't panic Alex, tired light is still hanging on among a couple of heretics, I think I'm one of them ? I will lead with a question for gary,
"what is the mean free path of a CMBR photon"? You may wish to read the link below first.
https://jumpshare.com/v/lCw7nbSN6P79vNU2zObh
In essence the standard candles are dimmed by dust attenuation, this gives rise to error in using any distant light for calibration. If you believe in "photons" and the wave function of the photon, then the wave function diverges 1/r with time and distance, and collapses for each dust particle en route. The dust particle gains momentum and warms. Three new photons are created , one at a slightly lower energy and frequency from the original and two RF photons, one narrow band as a result of the dust translating, and one blackbody as a result of warm dust. Total energy is conserved by the sum of all photons plus kinetic energy of the dust equaling the energy of the original photon. If you subsccribe to classical Maxwellian wave theory, as I do, then it is even simpler to come to this conclusion. The end result, non Doppler Redshift, optical attenuation and 2.7 K local blackbody radiation from dust.
Regards
Barry