Top quark mass $m_t$ at energy scales $\mu < m_t$? Edit - Maybe formulated differently:
Does it make sense to talk about the top mass at energies below $m_t$, although in all processes the corresponding energy scale is above $m_t$, because of the rest mass energy of the top quark?

In what sense is it or isn't it meaningful to speak about the top quark mass at energy scales below the top quark mass? Using an effective field theory approach, the top quark decouples at energies below the top quark mass and therefore has no influence on the mass running of the other fermions. Nevertheless, it is possible to compute how the top quark mass $m_t$ changes at energies below $m_t$, for example, because of the energy dependence of the gauge coupling constants. For example, in Updated Estimate of Running Quark Masses the authors compute

but write

We also illustrate the behavior of the heavy quark masses m q ( μ ) ( q = c,b,t ) in Fig. 3. Exactly speaking, the word “the running mass value m Q ( μ )” of a heavy quark Q at a lower energy scale μ than $μ = m_Q ( m_Q )$ loses the meaning. For example, the effective quark flavor number $n_q$ is three at μ = 1 GeV, so that the value of $m_t( μ )$  at $ μ = 1$  GeV has not the meaning. However, for reference, in Fig. 3, we have calculated the value of m Q ( μ ) ( Q = q N ) at μ n ≤ μ < μ n +1 ( n < N ) by using the relation m Q ( μ ) = c m Q R ( N ) ( μ ) [not m Q ( μ ) = c m Q R ( n ) ( μ )].

What do they mean by "loses the meaning"?
Similarly, the authors in Updated Values of Running Quark and Lepton Masses list in table II the top quark mass at various energy scales, for example $m_t(\mu= 2$ GeV) $ \approx 384.4$ GeV, which is considerably different from its value at the $Z$ scale $m_t(\mu= 91$ GeV) $ \approx 171.7$ GeV.
 A: Here thinking about effective theory approach and decoupling unnecessarily         complicates the issue. Just think about full standard model lagrangian.        Top quark mass is just another running/sliding parameter, depending on a
referent scale $\mu$. The choice of $\mu$ is free, and you are allowed             to take it $\mu = 2\,{\rm GeV}$, and $m_t(\mu)$ makes perfect sense even           then. Note, that $\mu$ is just a referent scale used to define                     renormalization procedure and, in principle, nothing should depend on              its value.                                                                         
In practice, you would keep the $\mu$ close to $m_t$ to avoid the appearance of large logarithms (which is the main reason to introduce running parameters in the first place), but you are not forbidden to define, use,                       and calculate with $m_t(2\,{\rm GeV}$.                                             
Also note that in quantum electrodynamics, you can work with                       renormalization scale $\mu = 0$ and electron mass is just fine with that.          
In practice, at low scales, you would work within effective theory,               decouple top quark, which then gets absorbed in other parameters,                  and doesn't exist any more in the theory, so then trivially its                    mass also becomes meaningless.
A: An answer can be found at page 176 of Srednicki's book: 
"If the particle mass is nonzero, this process stops at $µ ∼ m$. This is because the minimum value of $s$ is $4m^2$, and so the factor of $ln( \frac{s}{µ^2} )$ becomes an unwanted large $log$ for $µ ≪ m$. We should therefore not use values of $µ$ below $m$.
