Let $|f(x)|\le \dfrac{A}{1+x^2}$ for all $x$ and some $A$ (to ensure that $\int_{-\infty}^\infty f(x)\rm{d}x$ makes sense). I would like to show that $$g(z) = \int_{-\infty}^t f(x) e^{-2 \pi i z (x-t)}\rm{d}x$$ is continuous for all $z \in \mathbb{C}$ such that $\operatorname{Im}(z) \ge 0$.
I've tried the following: I'd like to show that $|g(z+h)-g(z)| \rightarrow 0$ as $h \rightarrow 0$ (where $h \in \mathbb{C}$). So I have \begin{align*}|g(z+h)-g(z)| &= \left|\int_{-\infty}^t f(x) e^{2 \pi i z (x-t)} (e^{-2\pi i h (x-t)}-1)\right|\rm{dx} \\ &\le\int_{-\infty}^t |f(x)| \cdot e^{2 \pi (\operatorname{Im}(z)) (x-t)} \cdot |e^{-2\pi i h (x-t)}-1|\rm{dx}\end{align*}
I'm not sure where to go from here though... any suggestions? I'm also worried about whether I need some sort of uniform convergence argument somewhere...
If the dominated convergence theorem is off limits, I think splitting the integral is the easiest way.
For $\operatorname{Im} w \geqslant 0$, we have
$$\left\lvert e^{-2\pi iw(x-t)}\right\rvert = e^{2\pi(x-t) \operatorname{Im}w}\leqslant 1,$$
hence
$$\int_{-\infty}^T \left\lvert f(x)\right\rvert \cdot \left\lvert e^{-2\pi i (z+h)(x-t)}- e^{-2\pi i z(x-t)} \right\rvert\,dx \leqslant 2\int_{-\infty}^T \lvert f(x)\rvert\,dx.\tag{1}$$
For a fixed $z$ with $\operatorname{Im}z \geqslant 0$, and a given $\varepsilon > 0$, choose $T < t$ so small that the right hand side of $(1)$ is smaller than $\varepsilon/2$.
Then, for $z+h$ in the upper half plane, we have
$$\begin{align} \lvert g(z+h) - g(z)\rvert &= \left\lvert\int_{-\infty}^t f(x) e^{-2\pi i z(x-t)}\left(e^{-2\pi i h(x-t)}-1 \right)\,dx\right\rvert\\ &\leqslant \int_{-\infty}^T \lvert f(x)\rvert \cdot \lvert e^{-2\pi i z(x-t)}\rvert\cdot \lvert e^{-2\pi i h(x-t)}-1\rvert\,dx\\ &\qquad + \int_T^t \lvert f(x)\rvert \cdot \lvert e^{-2\pi i z(x-t)}\rvert\cdot \lvert e^{-2\pi i h(x-t)}-1\rvert\,dx\\ &\leqslant 2\int_{-\infty}^T \lvert f(x)\rvert\,dx + A\cdot\int_T^t \lvert e^{-2\pi i h(x-t)}-1\rvert\,dx\\ &\leqslant \varepsilon/2 + A(t-T)\cdot \max_{T\leqslant x \leqslant t} \left\lvert e^{-2\pi i h(x-t)}-1\right\rvert\\ &\leqslant \varepsilon/2 + A(t-T)\cdot \left(e^{2\pi \lvert h\rvert(t-T)}-1\right). \end{align}$$
By the continuity of the exponential function, the last term can be made arbitrarily small by choosing a suitable bound for $\lvert h\rvert$. For $0 \leqslant r \leqslant 1$, we have $e^r - 1 \leqslant 2r$, so for
$$\lvert h\rvert \leqslant \frac{\varepsilon}{1 + \varepsilon + 2A(2\pi(t-T)^2)}$$
we have
$$A(t-T)\left(e^{2\pi\lvert h\rvert(t-T)}-1\right) \leqslant 2A\lvert h\rvert < \varepsilon/2.$$
Thus $g$ is continuous in $z$. Since $z$ was arbitrary, $g$ is continuous on the closed upper half plane.