Let's consider the function $$f: [0,2] \to \mathbb{R}, f(x) = \begin{cases}x^{-2} &\text{for}\;x\in(0,2] \\ 0 & \text{for} \;x =0 \end{cases}$$ Would this function be Riemann-integrable?
I'm having trouble thinking about this integral, can it be interpreted as $$\int_0^0 0 dx + \lim_{t\to 0} \int_t^2\frac{1}{x^2}dx$$ or would I need to take the formal approach to find a lower and upper step function and test wether $$\forall \epsilon >0 \exists\phi,\psi \in \mathcal{L}_{step}:\phi<f<\psi, \int(\psi - \phi)dx \leq \epsilon$$ How would I even define an upper step function? Wouldn't the first "step" have to be infinite to be larger then the whole function?
You say "How would I even define an upper step function? Wouldn't the first "step" have to be infinite to be larger then the whole function?" and that is exactly right: the function is not bounded and therefore not Riemann-integrable. It is Lebesgue-integrable, of course.