What's the importance of background field gauge? Recently I've read that background field gauge is very convenient for gauge theories, because it fixes the connection between normalization constants of gauge field and gauge coupling constant one. I don't understand corresponding Weinberg's explanation of this (volume 2, the same name paragraph, page 100). 
Can you give me the independent on Weinberg's one proof of that statement, or help to understand Weinberg statement from p. 100?
 A: I am also just first time reading it, and I am only a few months old in QFT, so there might be holes in my answer; but this is what I understand until now:

*

*The whole idea of using background field gauge was to keep the explicit gauge invariance. The way Weinberg defined the formal transformation, which on the shifted fields $A+A'$ basically acts like the regular gauge invariance, helps us have the regular gauge invariance on the background field. If unprimed are the background field, primed are the gauge field; see we have just the gauge transformations, under which the effective action is also invariant now:
\begin{align*}
    &\delta A_{\alpha}^{\mu} = \partial^\mu \epsilon_{\alpha} - C_{\alpha \beta \gamma} \epsilon_{\beta} A_{\gamma}^{'\mu}\\
    &\delta \psi = i t_{\alpha} \epsilon_{\alpha} \psi\\
    & \delta \omega_{\alpha} = - C_{\alpha \beta \gamma} \epsilon_{\beta}\omega_{\gamma}\\
    & \delta \omega^*_{\alpha} = - C_{\alpha \beta \gamma} \epsilon_{\beta}\omega^*_{\gamma}\\
\end{align*}

*Again, the effective action is just a function of these background fields, and has manifest gauge invariance because of the background field gauge and the formal gauge transformation we used. Now, because this gauge invariance is so stifling, it can only have a few types of renormalizable terms. Here, we are only writing out the possible terms with infinite coefficients :
\begin{align*}
    &\Gamma_{\infty} = \int d^4 x \mathcal{L}_{\infty}\\
    &\mathcal{L}_{\infty}  = -\frac{1}{4} L_A F_{\alpha \mu 
    \nu}F_{\alpha}^{\mu \nu} - L_{\psi} \bar{\psi}\gamma^\mu\bar{D}_{\mu}\psi - m L_m \bar{\psi}\psi - L_{\omega} (\bar{D}_\mu \omega^*_\alpha)(\bar{D}^{\mu}\omega_{\alpha})
\end{align*}

*Next, we need to evaluate the effective potential, the zeroth order term has just the classical Lagrangian with background fields (see chapter 16 for this as well); now if we renormalize just the background fields the right way, for example for the field  with a factor $\sqrt{1+L_A}$, we find that the coupling constant has the same renormalization factor, i.e. $g^R = g(1+L_A)^{-1/2}$
