Understanding a dominated convergence argument from Rudin

Question

The reference is the beginning of section 19.1 from RCA Rudin. Please check it if you need context.

The part I got confused is that:

If $\mathrm{Im~} z > \delta > 0$, $\mathrm{Im~} z_n > \delta$, and $z_n \to z$, the dominated convergence theorem shows that $$\lim_{n \to \infty} \int_0^\infty|\exp (itz_n) - \exp (itz)|^2 dt = 0$$ because the integrand is bounded by the $L^1$-function $4 \exp (-2 \delta t)$ and tends to $0$ for every $t > 0$.

I don't understand. The dominated convergence theorem said if $f_n$ converges to $f$ pointwise and $f_n$ is bounded by an $L^1$-function $g$ then $f_n \to f$ in $L^1$ sense. Or $L^p$ condition yields $L^p$ convergence. How could the book get something squared? And why did the book bound the whole integrand?

Answer 1

Apply the dominated convergence theorem to $f_n(t)= \vert e^{itz_n}-e^{itz} \vert ^2$ and $f(t)=0$.

Answer 2

I think that is the $L^2$ limit, per your comment about $L^p$ condition implies $L^p$ convergence.