How is a continuous spectrum for the DTFT possible? So we that a complex sinusoid of the form $e^{j\omega_0n}$ is periodic over $N=2\pi/\omega_0$ only if $\omega_0$ is a rational multiple of $\pi$, otherwise the exponential is not periodic. (see EDIT!)

Then we got to the DTFT, specifically the synthesis equation: 

$$x[n]=\frac{1}{2\pi}\int_{2\pi}X\left(e^{j\omega}\right)e^{j\omega n}d\omega$$

Here $X\left(e^{j\omega}\right)$ if a function that's continuous over $\omega$, and $X\left(e^{j\omega_0}\right)$ is how much a complex sinusoid $e^{j\omega_0n}$ contributes to $x[n]$. 

But that confuses me, because I feel like not all sinusoids should be able to contribute to $x[n]$ since not all complex sinusoids are periodic. Or are aperiodic sinusoids actually the key to the DTFT?

EDIT: I should clarify that in that first bit where I talk about complex sinusoids being periodic, I'm talking about discrete complex sinusoids, $x[n] = e^{j\omega_0n}$ where $n\in\mathbb{Z}$. For $x[n]$ to be considered periodic there must be some $N\in\mathbb{Z} \mid x[n+N]=x[n]$. For such an $N$ to exist, $\omega_0$ must be a rational multiple of $pi$.

My confusion comes from the fact that the synthesis equation includes values of $\omega$ that are not rational multiples of $\pi$, so aperiodic discrete sinusoids are contributing to synthesizing $x[n]$.
The reason the DTFT has values at frequencies that result in discrete aperiodic sinusoids is because we are interested in representing all possible discrete time signals including (linear combinations of) discrete aperiodic sinusoids. Let's say that we have a signal $x[n] = e^{j \omega_0 n}$ where $\omega_0$ is not a rational multiple of $\pi$. Would you like this signal to have a DTFT? Of course. It would not have one if we did not allow the existence of the DTFT at all frequencies (including irrational multiples of $\pi$). 

Put another way, periodicity is not a requirement for the existence of the DTFT of a signal.