Is mass an inherent property? Suppose I have an electronic weighing machine placed in a uniform gravitational field. Now I put a mass above it and register the reading. Now I give the system (mass + machine) an impulse so that it travels at $0.99c$. Will the reading on the scales change and why?
 A: In short the answer is no - the measurement of mass depends on your frame. Here's a (somewhat lengthy) explanation.
First let's be clear about the setup. We'll make the following assumptions


*

*an object $A$ is at rest in the frame of an observer $O$

*$O$ measures a uniform gravitational field in some direction of strength $g$


Now $O$ can measure the mass of the object $A$ in two ways namely


*

*put $A$ on a set of scales at rest and determine the weight then divide by the field strength $g$

*apply a force to $A$ and measure its acceleration then find the mass by $F = ma$


These two methods give the same answer in all experiments ever done, and determine the rest mass of $A$.
Now suppose that relative to a second observer $O'$ the whole setup containing $O$ and $A$ is moving at $0.99c$. This doesn't affect what $O$ himself measures, since there's no physical meaning in relative inertial motion according to special relativity.
But we really want to know $O'$ would measure. So suppose $O'$ had his own set of scales which he put under $A$ at the instant he passed $O$. 


*

*first note that $O'$ would measure the same field strength $g$ as $O$ because the field is uniform and the laws of physics are the same in all inertial frames

*$O'$ would get a different weight reading on his scales and therefore conclude that $A$ had a different mass to that measured by $O$


But why should $O$ get a different mass? We can do a simple "proof" if you accept Einstein's axiom
$$E = mc^2$$
where $E$ is the total energy of $A$ in the frame of $O'$ and $m$ is the mass measured by $O'$. The argument goes like this.


*

*The mass measured by $O'$ is thus $m = E/c^2$

*From the perspective of $O'$ $A$ is moving very fast so has more kinetic energy that it seems to $O$

*Hence the energy measured by $O'$ is larger than that measured by $O$

*So $O'$ measures a larger mass than $O$


The mass measured by $O'$ is called the relativistic mass. 
In reality nobody really works with the relativistic mass because people can't agree on it. Everyone just uses the rest mass because it's the same for all observers by definition! Indeed if $O'$ took $A$ with him at $0.99c$ and weighed it he'd get the same answer as $O$ got before.
A caveat: things aren't so simple when you think about general relativity. It's difficult to precisely define local rest mass for particles, because the fundamental conserved quantity is the density of energy and momentum through spacetime. As a result $O$ and $O'$ might even disagree about the rest mass of $A$. 
The heuristic reasoning is as follows. In GR there is a constant exchange of energy and momentum through the universe. The net rate of this exchange depends on your frame of reference. So changing frames changes your observations about mass in the universe.
References
An experiment to measure relativistic mass
Matt Strassler on Rest Mass
Harvey Reall on Energy in GR (see sec 5.2)
