Why is the sunset not bluer My question is a duplicate of this; Clarification on Rayleigh scattering causing various sky colors.
The accepted answer from the link above says that at sunset the scattering occurs farther away and does not reach the observer, which is unsatisfactory and vague to me. (not even sure if it's the right and correct answer)
It doesn't quite make sense to me that at noon, we see scattered blue light whereas, at sunset, we see the sunlight itself minus the scattered blue light. At noon or at sunset, we are not seeing the sun, but the scattered sunlight indirectly. The logic should apply both cases equally and should imply that at sunset we see bluer sky than at noon because of more scattering.
Please let me know if this duplicate question was unnecessary because the accepted answer from the link above was enough to answer the question.
 A: Considering 11 km as the thickness of the atmosphere that scatters the blue light, when the sun is just at the horizon, the length travelled by the sun light is:
$$l = \sqrt{(6371+11)^2 - 6371^2} = 374 km.$$ 
When that rays come to the place where we are looking at the horizon, much of the blue components were already scattered along the long journey through the atmosphere.
We see light deprived from bluish frequencies, what look reddish to our eyes. 
A: The Rayleigh scattering cross-section from nitrogen molecules is around $1.6\times 10^{-30}$ m$^2$ at 400 nm (blue light). At sea level, there are about $2\times 10^{25}$ molecules per cubic metre. The mean free path of a blue photon is therefore 31 km. Since the Earth's atmosphere has a scale height of around 8 km, then the vast majority of photons travelling from the Sun at zenith, directly towards the ground, make it to sea level. Hence the colour of the Sun is changed very little when it is overhead.
It is also the case that because the mean free path of a blue photon is greater than the atmospheric scale height, that most photons that are scattered, are only scattered once before reaching the ground, because the location where they are scattered is closer than their mean free path to the observer. 
On the contrary, when we look towards the horizon, the path length through the atmosphere is drastically increased. When the Sun is 5 degrees above the horizon (about 20-30 minutes before sunset) the "airmass" that photons must pass through is increased by a factor of 10. So now there are roughly 3 mean free paths of nitrogen in the way of direct photons from the Sun and there is therefore almost no chance of a blue photon traversing this without being scattered. Further, any blue photons that are scattered through a small angle are also unlikely to make it to the observer, because they will be multiply scattered. The result is that blue photons are almost absent from a direct line towards the Sun.
A: From one of your comments to other answers:

I think the sunlight doesn't lose its blue that much on the way to the point where red scattering dominates.

Why wouldn't it be able to?
Rayleigh scattering has cross section proportional to $\lambda^{-4}$. The spectrum of light illuminating the volume scattering it gets multiplied by $\lambda^{-4}$, which indeed makes smaller wavelengths amplified more than larger ones. This indeed makes the scattered light, when seen immediately after scattering, bluer.
But on the other hand, this same scattering mechanism removes corresponding amount of power from the light that hasn't gotten scattered and continues propagating forwards. What we now have is this factor of the scattering cross section getting into the exponent of the Beer-Lambert law. In the limit of large distance $d$, the factor $\lambda^{-4}\exp(-d \lambda^{-4})$ has a reddening effect, not bluing. Here the distance you have to take into account is the sum of 1) distance traversed before scattering and 2) distance traversed to you after scattering.
Now, even if you take into account double scattering, which would give you a second factor of $\lambda^{-4}$, you'll still get $(\lambda^{-4})^2\exp(-d_\Sigma \lambda^{-4})$ where $d_\Sigma$ is the sum of distances traversed by the sunlight 1) before first scattering, 2) between first and second scatterings and 3) after second scattering until it got to you. This is even larger distance, and exponential function in the above mentioned expression, again, is much more selective by wavelength than the simple power in the factor it's multiplied by.
The result of this reddening, apart from the orange solar disk, is the phenomenon of Belt of Venus — the area of the sky on the opposite of the solar azimuth, which has purple-red color.
There's also a peculiarity of the Earth atmosphere that fools humans into thinking that the blue sky at day and blue sky at twilight have the same reason. Actually, if the atmosphere had no ozone, twilights also wouldn't even be as blue as they are. Beer-Lambert law would have much greater effect. On the Earth the twilight is blue because the sunlight, traveling in the upper parts of the atmosphere, gets absorbed in the red part of the spectrum by the Chappuis absorption band of ozone. If not the ozone layer, twilights would actually have sandy-brown color.
A: When you look at the sun overhead (not advisable) you see the white light from the sun with a little bit of the blue scattered horizontally.  The blue sky is blue light scattered from sunbeams going elsewhere.  When the sun is near your horizon, its light passes through a much greater distance of dense atmosphere, and most of the blue is lost from the beam to give other people a blue sky.
