Denote $I = (1, \infty)$ and define the function $f : \mathbb R^+ \to \mathbb R$ as $$ f(t) = \left\{ \begin{aligned} t, && t \in \mathbb N, \\ 0, && t \notin \mathbb N\,. \end{aligned} \right. $$ is the function a member of $L^\infty(I)$, $L^2(I)$ and/or $L^3(I)$?
An incomplete attempt
As per the definition of the essential supremum, we can ignore the points $t \in \mathbb N$ from the domain of $f$, so that $\Vert f \Vert_{L^\infty} = \mathrm{ess\,sup}_{t \in I} |f(t)| = 0 < \infty$, so $f \in L^\infty(1, \infty)$. However, I'm stuck in the case of the other two $p$-norms, which were defined as $$ \Vert f \Vert_{L^p} = \left( \int_I | f |^p \, \mathrm d t \right)^{1/p}, $$ where the integral is the Lebesgue integral, not the Riemann integral.
Then for example $$ \Vert f \Vert_{L^2} = \left( \int_I | f |^2 \, \mathrm d t \right)^{1/2}, $$ but how do I evaluate this? Do I simply conclude that on each interval $(t, t+1) \subset (1, \infty)$, the infimum of $f$ is $0$, so the integral is $0$ as well (as it is defined as the supremum of the lower Lebesgue sums on $I$)?
Your definition of the $L^2$ and $L^3$ norms is wrong, you should be squaring and cubing inside the integral. But the idea is the same, the support of $f$ is a null set so from the point of view of the $L^p$ spaces it is the zero function.
How exactly you see this directly from the definition depends on which definition you use. If you use the supremum of Lebesgue integrals of nonnegative simple functions below $f$ (which works for nonnegative $f$) then you do indeed get this result, since all simple functions below $f$ are identically zero.