Mathematical Reflections 2017, Issue 6 - Problem U427

Problem:
Let $f:\mathbb{R}^2 \to \mathbb{R}$ be a function defined by $f(x,y)=\mathbf{1}_{(0,1/y)}(x)\cdot\mathbf{1}_{(0,1)}(y)\cdot y$, where $\mathbf{1}$ is the characteristic function. Evaluate
$$\int_{\mathbb{R}^2} f(x,y) \ dx \ dy.$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Since $f$ is clearly non-negative, by Tonelli's Theorem we can evaluate the double integral as iterated integrals, i.e.
$$\int_{\mathbb{R}^2} f(x,y) \ dx \ dy=\int_{\mathbb{R}} \left(\int_{\mathbb{R}} f(x,y) \ dy\right) \ dx.$$
Observe that
$$\renewcommand{\arraystretch}{2}\begin{array}{lll} \displaystyle \int_{\mathbb{R}} f(x,y) \ dy&=& \displaystyle \int_{\mathbb{R}} \mathbf{1}_{(0,1/y)}(x)\cdot\mathbf{1}_{(0,1)}(y)\cdot y \ dy\\&=& \displaystyle \int_{\mathbb{R}} \mathbf{1}_{(0,1/x)}(y)\cdot\mathbf{1}_{(0,1)}(y)\cdot y \ dy\\&=& \displaystyle \int_{\mathbb{R}} \mathbf{1}_{(0,\min(1,1/x))}(y)\cdot y \ dy. \end{array}$$
Since $$\mathbf{1}_{(0,\min(1,1/x))}(y)=\begin{cases} \mathbf{1}_{(0,1)}(y) & \textrm{if } 0 < x \leq 1 \\ \mathbf{1}_{(0,1/x)}(y) & \textrm{if } x > 1, \end{cases}$$
it follows that $$\int_{\mathbb{R}} \mathbf{1}_{(0,\min(1,1/x))}(y)\cdot y \ dy = \begin{cases} 1/2 & \textrm{if } 0 < x \leq 1 \\ 1/(2x^2) & \textrm{if } x > 1. \end{cases}.$$
So, $$\int_{\mathbb{R}^2} f(x,y) \ dx \ dy=\int_0^1 \dfrac{1}{2} \ dx+\int_{1}^{\infty} \dfrac{1}{2x^2} \ dx=1.$$