Analog TV signals have one sideband cut off to reduce their (very wide) bandwidth and save frequency space in the ether. The lower sideband is cut off, but not completely, the lower 1.25 MHz is kept resulting in a vestigial sideband signal. I've read several explanations on why this vestigial sideband is kept, but they don't fully make sense to me.
Fully removing the lower sideband requires a filter with a very sharp edge because the signal contains significant low frequency information. Such a sharp filter is not practical, so as a compromise part of the lower sideband is kept. This makes total sense, but doesn't explain why you would want to keep a full 1.25 MHz of bandwidth. Surely the bandwidth filters at the time analog TV was developed could do better than that?
"The video signal has significant low-frequency content (average brightness) and has rectangular synchronizing pulses. The engineering compromise is vestigial-sideband transmission." is what Wikipedia says. That makes no sense to me. All the information is present in only one sideband, so as long as one sideband is preserved without damage the original signal (including synchronization pulses, etc) can be recovered.
It was desirable for TV sets to be able to use an envelope detector, which is cheaper than an SSB demodulator. That reasoning makes sense, but as far as I know envelope detectors only work for AM signals, not VSB. So I have no idea if this reasoning is just random internet nonsense or if there is some substance to this.
One reason I can imagine that makes it fit together is if TV sets use an envelope detector to extract the timing pulses and an SSB demodulator for the luminance signal, but I have not been able to find any corroboration of this idea.
So, what was the actual reason analog TV uses VSB with a full 1.25 MHz of lower sideband?