I'm reading about Muckenhaupt $A_p$ weights in Grafakos' book Classical Fourier Analysis and the first proposition he gives is partially unclear (and stated as straightforward, so no solution given). There are three statements that I'm trying to understand; I've only worked 1 through though. The general Muckenhaupt weight definition is $$ [w]_{A_p} = sup_{I \in Q} \big\{ (\frac{1}{|I|} \int_I w(x)dx)(\frac{1}{|I|} \int_I w(x)^{\frac{-1}{p-1}} )^{p-1}\big\}.$$
Grafakos claims for a Muckenhaupt weight $w \in A_p$, that $ [\delta^{\lambda}]_{A_p} = [\tau^z(w)]_{A_p}=[\lambda w]_{A_p}=[w]_{A_p}$, with $[\delta^{\lambda}]_{A_p} = (\delta^{\lambda}w)(x) = w(\lambda x),$ $[\tau^z(w)]_{A_p} = \tau^z (w)(x) = w(x-z), z \in \mathbb{R^n},$ and finally $[\lambda w]_{A_p} = \lambda w(x)$.
I understand the last function, and how it equals the original $[w]_{A_p}$ function; it cancels very cleanly. However, for the other (first) two they change the input of the function, not just the scale of the function; so how can that be the same as the original?
How can that be shown to be equivalent when we're dealing with a function's input nested in an integral average? Any help would be really appreciated, this seems like it's a lot simpler than I'm making out to be.