Let $f:\mathbb R→ \mathbb R$ be given by $f(x)=⌊x^{3}⌋$ To explain why the function $f$ is integrable over any closed interval [a, b].
My answer currently would be: Given the function $f$ effectively maps to a series of integers in a discrete manner (and $f$ is bounded over closed interval), the graph can be partitioned into a series of constant functions. Given that in general we know that the set of upper Darboux sums and the set of lower Darboux sums are equivalent on constant functions, it is obvious that this function lies in the set of Riemann integrable functions over the interval $[a,b]$.
Am I missing anything? comments say I should be more precise
Attempt 2
Let us begin by describing our partition given the interval $[a,b]$. 
Let |{$z \in \mathbb Z :z \in  [⌊a^3⌋,⌊b^3⌋]$}| $= n\lt\infty$. Consider the evenly distribution partition of $[a,b]$ defined as $P=${$x_i$ : $x_i = \frac{b-a}{k}$ where $k\le n$}.
But this would be wrong I realised as I can immediately think of something that won't work.
(I'll continue working on this)
 A: I would try something along the lines:
$f$ is Riemann integrable in $[a,b]$ if it's bounded in $[a, b]$ and $\exists L\in\mathbb{R}$ if $\forall \epsilon >0 \exists \delta_f(\epsilon)$ such that
given a partiton $\mathbf{P}$ of [a, b], if $\lVert P \rVert<\delta_f(\epsilon) \Rightarrow |S(f,\mathbf{P})-L|<\epsilon$ 
It's easy to see $f$ is bounded in any interval $[a, b]$, because $f$ is monotonic.
$f$ is discontinuous for $x\in \lbrace x:x=\lfloor x^3 \rfloor \rbrace $. Then f can be written as the sum of several functions:
Let $k\in\mathbb{Z}$
Let $f_k = \begin{cases}
k &k=\lfloor x^3 \rfloor\\
0 &otherwise
\end{cases}$
in particular $f_k\neq 0$ only in $[\sqrt[3]{k}, \sqrt[3]{k}+1)$
$f=\underset{k\in \mathbb{Z}}{\sum}f_k$
$f \cap[a,b] = [a, b]\cap \underset{k\in \mathbb{Z}}{\sum}f_k = [a, b]\cap\underset{k\in \mathbb{Z}\cap [f(a), f(b)]}{\sum}f_k$
Now it's obvious that your function is Riemann integrable using the partition
$\mathbf{P_0}=\lbrace a < \lfloor a^3 + 1\rfloor^{1/3} < \lfloor a^3 + 2\rfloor^{1/3} < \dots < b \rbrace$
Then calculate S for that partition, taking into account that in each interval corresponding to a k, $f$ would be $f_k$ and that's constant
Then to make $\lVert P \rVert$ as small as you want adding more points, it won't change the fact that the sum would reduce to a constant function $f_k$ between the points of $\mathbf{P_0}$ and those are included in any other $\mathbf{P}$
