In this proof for CLT, the author said "it is no restriction to assume that $\mu=0, \sigma=1$". This is not clear to me. Can anyone share your thoughts?
My attempt
Note that \begin{align} \varphi_{\frac{S_n-n\mu}{\sigma\sqrt{n}}}(t) &= E ( e^{it \frac{S_n-n\mu}{\sigma\sqrt{n}}} ) \\ &= \int_{-\infty}^{\infty} e^{it \frac{s_n-n\mu}{\sigma\sqrt{n}}} \cdot f_{S_n}(s_n) ~ds_n, \end{align} and, \begin{align} \varphi_{\frac{S_n}{\sqrt{n}}}(t) &= E ( e^{it \frac{S_n}{\sqrt{n}}} ) \\ &= \int_{-\infty}^{\infty} e^{it \frac{s_n}{\sqrt{n}}} \cdot f_{S_n}(s_n) ~ds_n. \end{align}
These two are NOT equal.
My thoughts
I think the equality in blue circle does not hold quantitatively in general. It holds based on some explanation and understanding.
Following Andrew's comment, I will give a proof or explanation.
Proof
We want to know the distribution of \begin{align} \frac{(\sum_i X_i) -n\mu}{\sigma\sqrt{n}}. \end{align}
However, \begin{align} \frac{(\sum_i X_i) -n\mu}{\sigma\sqrt{n}} = \frac{\sum_i \frac{X_i - \mu}{\sigma}}{\sqrt{n}} = \frac{\sum_i Y_i}{\sqrt{n}}, ~~~~~~~~~~~~(1) \end{align} where \begin{align} E(Y_i) &= 0 \\ Var(Y_i) &= 1. \end{align}
That is, the LHS and RHS of Eq (1) have the same distribution. That is why we can just focus on $Y_1, ..., Y_n$ instead.