Prove $\int\int_{Q} \left\lfloor x + y\right\rfloor\mathrm{d}x\,\mathrm{d}y$ exists where $Q \equiv \left[0,2\right] \times \left[0,2\right]$.
I was trying to find some boundings for the function but IDK how to get to $\mbox{Sup}\ S = \mbox{Inf}\ T$ in order to prove the integrablity of the function.
On
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{2}\int_{0}^{2}\left\lfloor x + y\right\rfloor \dd x\,\dd y} = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \left\lfloor x + y\right\rfloor\bracks{0 < x < 2}\bracks{0 < y < 2} \dd x\,\dd y \\[5mm] = &\ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \left\lfloor x\right\rfloor\bracks{0 < x - y < 2}\bracks{0 < y < 2} \dd x\,\dd y \\[5mm] = &\ \int_{-\infty}^{\infty}\left\lfloor x\right\rfloor\int_{-\infty}^{\infty} \bracks{x - 2 < y < x}\bracks{0 < y < 2}\dd y\,\dd x \\[5mm] = &\ \int_{-\infty}^{\infty}\left\lfloor x\right\rfloor\braces{% \bracks{0 < x < 2}\int_{0}^{x}\dd y + \bracks{0 < x - 2 < 2}\int_{x - 2}^{2}\dd y}\,\dd x \\[5mm] = &\ \int_{-\infty}^{\infty}\left\lfloor x\right\rfloor\braces{% \bracks{0 < x < 2}x + \bracks{2 < x < 4}\pars{4 - x}}\,\dd x \\[5mm] = &\ \int_{0}^{2}\left\lfloor x\right\rfloor x\,\dd x + \int_{2}^{4}\left\lfloor x\right\rfloor \pars{4 - x}x\,\dd x \\[5mm] = &\ \underbrace{\int_{1}^{2}x\,\dd x}_{\ds{3 \over 2}}\ +\ \underbrace{\int_{2}^{3}2\pars{4 - x}\,\dd x}_{\ds{3}}\ +\ \underbrace{\int_{3}^{4}3\pars{4 - x}\,\dd x}_{\ds{3 \over 2}}\ =\ \bbx{\large 6} \\ & \end{align}
Here is a suggestion that will help you calculate the integral. The value of $\lfloor{x+y}\rfloor$ is constant in each of the colored regions I depicted in the hyperlink.