Why is a liquid-vapour mixture at thermal equilibrium called saturated? If we have a liquid-vapour mixture at thermal equilibrium, apparently this means the mixture is saturated.
How?
From what I understand, saturated means it is on the verge of converting to either liquid or gas. But it also apparently means that the thermodynamic properties of the steam and water are the same.
How does the fact the system is in thermal equilibrium tell us this?
 A: There is some variation in the use of the terms, but in general:
The term "saturated" refers to vapor in equilibrium with liquid at or above the normal boiling point (boiling point at one atmosphere, in the case of water). Adding or removing heat heat increases the vapor component or increases the liquid component, respectively.
The term saturated vapor refers to 100% vapor on the verge of either being converting to liquid (if heat removed) or superheated vapor (if heat added).
The term saturated liquid refers to 100% liquid being on the verge of either converting to vapor (if heat added), or lowering in temperature (if heat removed).
Hope this helps.
A: 
If we have a liquid-vapour mixture at thermal equilibrium, apparently this means the mixture is saturated.

This is correct if you remove "thermal," which implies only that the temperatures everywhere are equal. A mixture of liquid with its own vapor at half the saturation pressure, for example, could easily be at thermal equilibrium, with uniform system temperature. However, it's not at complete equilibrium because it's not in chemical equilibrium; the chemical potentials of the two phases aren't equal.
The chemical potential of the gas $\mu_\mathrm{G}$ and partial pressure $p$ are related through
$$\mu_\mathrm{G} =\mu^\ominus_\mathrm{G}+RT\ln\frac{\phi p}{p^\ominus},$$
where $\mu^\ominus$ is the chemical potential at a defined set of standard conditions, $R$ is the gas constant, $T$ is temperature, $\phi$ is the fugacity coefficient (approximately 1 for dilute vapor), and $p^\ominus$ is the pressure at the defined conditions.
At chemical equilibrium, the chemical potentials of the gas and solid must be equal:
$$\mu_\mathrm{S}=\mu_\mathrm{G};$$$$\mu_\mathrm{S}^\ominus= \mu^\ominus_\mathrm{G}+RT\ln\frac{\phi p}{p^\ominus}.$$
(We don't see an $RT$ term for pure condensed matter because its activity is always taken to be 1.)
From this, we find that the partial pressure $p$ evolves to an equilibrium value governed by the difference $\mu_\mathrm{S}^\ominus-\mu_\mathrm{G}^\ominus$ between the solid and gas phases. Since the chemical potential is the partial molar Gibbs free energy and since changes in the Gibbs free energy $\Delta G=\Delta H-T\Delta S$, where $H$ is enthalpy and $S$ is entropy, the equilibrium vapor pressure depends on the change in enthalpy (also known as the latent heat) and change in entropy that occur upon a phase change at the temperature of interest.
The process of evaporation/condensation tends to heat or cool the participant phases; heat transfer eliminates any resulting temperature differences, providing us ultimately with complete (thermal and chemical) equilibrium.
A: 
If we have a liquid-vapour mixture at thermal equilibrium, apparently
this means the mixture is saturated. How?

If the vapour containing air in the water vapour mix is not saturated, it will accumulate vapour from the water. This stagnates only to a standstill, if the air is saturated. Thus, the presence of water, in an equilibrium state, would indicate that the air is saturated.
A: Sorry for my poor english.
Saturating vapor is opposed to dry vapor. Imagine increasing the amount of water vapor in a container at constant temperature (and slowly to be in equilibrium). The pressure of this water vapor increases gradually. As long as there is no liquid, it is dry vapor. There will come a time when you will no longer be able to increase this pressure: the added water vapor turns into liquid. The steam is then saturated. This vapor pressure cannot be exceeded at the imposed temperature.
When the vapor is saturated, the chemical potentials (molar free enthalpy) of the liquid and the vapor are equal. This is necessary because otherwise the system would tend to evolve towards the phase (all liquid or all vapor) of the lowest chemical potential at this pressure and temperature.
But other quantities such as density, molar entropy.... are different. They tend to approach when the temperature approaches the critical point.
