I'm working through a statistical theory text and have this proof that if $X$ is a normally distributed random variable then the random variable $Z = (X - \mu)/\sigma$ has a n(0,1) distribution. The proof is as follows:
$$P(Z\leq z) = P(\frac{X-\mu}{\sigma}\leq z)$$ $$= P(X \leq z\sigma + \mu)$$ $$=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{z\sigma + \mu}e^{-(x-\mu)^2/(2\sigma^2)}dx$$ $$=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z}e^{-t^2/2}dt$$
(The book uses t instead of u for the u-substitution, I'm guessing because it would look too similar to $\mu$)
I understand all of the moves in this, except the removal of the sigma constant outside of the integral going from steps 3 to 4. The substitution is $t=\frac{x-\mu}{\sigma}$, which makes sense to me within the integrand and in changing the upper parameter of the integral, but I can't figure out at all why or how it would cancel out the sigma outside of the integral. I appreciate any help.
One may notice that the substitution $t=\frac{x-\mu}{\sigma}$ gives $dt=\frac{dx}{\sigma}$ then $$\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{z\sigma + \mu}e^{-(x-\mu)^2/(2\sigma^2)}dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z\sigma + \mu}e^{-(x-\mu)^2/(2\sigma^2)}\frac{dx}{\sigma}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z}e^{-t^2/2}dt.$$