Suppose $f$ is continuous and of moderate decrease. i.e.
$$|f(x)| \le \frac C{1+x^2},\forall x \in \mathbb R, C>0$$
For fixed $t \in \mathbb R$: $$A(z)=\int_{- \infty}^{t} f(x)e^{-2\pi iz(x-t)}dx$$
and $$B(z)=- \int_{t}^{\infty} f(x)e^{-2\pi iz(x-t)}dx$$
The aim is to show that $A(\xi) = B(\xi)$, $\forall \xi \in \mathbb R$
Using this result, the second aim is to prove that the function:
$$F=\begin{cases} A, & \text{in the closed upper half plane} \\ B, & \text{in the lower half plane} \\ \end{cases}$$
is entire and bounded and therefore constant.
Honestly, I've been working at the this problem for a while now and I think I can prove $|A(\xi)| = |B(\xi)|$ but that's not the same so I'm not sure...
This is not true. Take $$ f(x)=\begin{cases} 1+\frac{x}{2} & \text{ for }\, -2 \le x< 0\\ 1-x & \text{ for }\, 0\le x\le 1\\ 0 & \text{ otherwise } \end{cases}. $$ It is clear that for $C>0$ large enough $$ |f(x)|\le \frac{C}{1+x^2} \quad \forall x \in \mathbb{R}. $$ For $t=0$ we have \begin{eqnarray} A(z)&=&\int_{-2}^0f(x)e^{-2\pi izx}\,dx=\int_{-2}^0(1+\frac{x}{2})e^{-2\pi izx}\,dx=\frac{2+x}{-4\pi iz}e^{-2\pi izx}\Big|_{-2}^0+\frac{1}{4\pi iz}\int_{-2}^0e^{-2\pi izx}\,dx\\ &=&\frac{i}{2\pi z}+\frac{1-e^{2\pi iz}}{8\pi^2z^2},\\ B(z)&=&-\int_0^1f(x)e^{-2\pi izx}\,dx=-\int_0^1(1-x)e^{-2\pi izx}\,dx=-\frac{1-x}{-2\pi iz}e^{-2\pi izx}\Big|_0^1+\frac{1}{2\pi iz}\int_0^1e^{-2\pi izx}\,dx\\ &=&\frac{i}{2\pi z}-\frac{1-e^{-2\pi iz}}{4\pi^2z^2}. \end{eqnarray} You can also show that if $t \in (-1,1)$ then you always have $A\not\equiv B$.