For $\lambda\in \mathbb{R} $, consider the complex exponential $e_{\lambda}: \mathbb{R} \to \mathbb{C}$
$$e_{\lambda}(t) : = e^{2\pi i t \lambda } $$ For a set $\Lambda \subset \mathbb{R}$, let $E(\Lambda) := \{e_{\lambda}| \lambda \in \Lambda \} $. I want to show that $E(\mathbb{Z} \cup (\mathbb{Z} + \frac{1}{4}) ) $ is an orthogonal basis for $L^2(S)$ where $S$ is the union of open intervals $S = (0,1) \cup (2,3)$. This requires proving orthogonality and completeness.
Orthogonality isn't too difficult. It's clear that $E(\mathbb{Z})$ and $E(\mathbb{Z} + \frac{1}{4})$ are orthogonal in $L^2(S)$. Let $m,n \in \mathbb{Z}$. We need to show that that $\langle e_m, e_{n+1/4} \rangle_{L^2(S)} = 0 $.
\begin{align} \int_S e^{2\pi i mt} e^{-2\pi i (n +1/4)t} dt = & \int_0^1 e^{2\pi i (m-n - 1/4)t} dt + \int_2^3 e^{2\pi i ( m - n - 1/4)t } dt \\ = & \frac{1}{2\pi i(m-n - 1/4)} \bigg( e^{2\pi i (m-n - 1/4)} - 1 + e^{6\pi i (m-n - \frac{1}{4})} - e^{4\pi i (m-n - \frac{1}{4})}\bigg) \\ = & \frac{1}{2\pi i(m-n - 1/4)}\ ( e^{-\pi i/2} - 1 + e^{-3\pi i/2} - e^{-\pi i} ) \end{align}
The term in the paranthesis is equal to $ -i - 1 + i + 1 = 0$ so orthogonality is satisfied.
I have no idea how to approach completeness for $L^2(S)$. I'm not even totally sure it's true, but it seems like it is.