Fourier Transform Units It is documented that 'one' of the units of the Fourier Transform [of $x(t)$ volt] is volt per Hz. That is $X(\omega)$ components will have units of volt per Hz, where $\omega$ is the angular frequency (along the $x$-axis). 

For example, the Fourier Transform of $\cos(\omega_0 t)$ volt has two components - one at $\omega = \omega_0$ and one at $\omega = -\omega_0$, each assigned a 'value' of $\pi$. So it is assumed that the values are '$\pi$ volt per Hz'.

Now, when the plot is adjusted to become a function of cyclic frequency $f$ (in Hz), which means plotting $X(f)$ versus $f$, then the Fourier Transform components of $\cos( 2\pi f_0 t)$ become '1/2 or 0.5' instead of '$\pi$'.

My question is --- why isn't the '$\pi$ volt per hz' preserved when translating between $X(\omega)$ and $X(f)$?

According to tables showing Fourier Transform pairs, the Fourier Transform of  $\cos( 2\pi f_0 t)$ volt has two components -- each with a value of '1/2 or 0.5'. For these values, does anyone know what are their units? I had assumed that the values for each component would have been 'preserved' (to be $\pi$ volt per Hz), regardless of whether we do the plot as a function of '$\omega_0$' rad/s or a function of '$f_0$' Hz. Thanks all.
For a given signal $g(t)$ with Fourier transform $G(\omega)$:

$$g(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} G(\omega)e^{i\omega t}\textrm{d}\omega$$

If you change variables such that $\omega=2\pi f$

$$g(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} G(2\pi f)e^{i2\pi ft}\textrm{d}(2\pi f)=\int_{-\infty}^{\infty} G(2\pi f)e^{i2\pi ft}\textrm{d}f$$

So if you present your Fourier transform as a function of $f$ instead of $\omega$, your Fourier pair of $g(t)$ will now be $\hat{G}(f)=\frac{G(2\pi f)}{2\pi}$ and:

$$g(t) =\int_{-\infty}^{\infty} \hat{G}(f)e^{i2\pi ft}\textrm{d}f$$

Note that with this definition, there is a scaling of $2\pi$ in comparison with the first equation I wrote. So now, for example, the Fourier transform of a cosine will consist of two deltas not with amplitude $\pi$, but with amplitude $\frac{\pi}{2\pi}=\frac12$, as correctly stated in your question.