Let $f:[0,1]\times [0,1] \subseteq \mathbb{R}^2 \rightarrow \mathbb{R}$ defined as : \begin{equation} \label{eq:aqui-le-mostramos-como-hacerle-la-llave-grande} f(x,y) = \left\{ \begin{array}{ll} 1 & \mathrm{if\ } x ~is~rational \\ 2y & \mathrm{if\ } x~is~irrational \end{array} \right. \end{equation}
I have to show that $f$ is continuous on $[1,0]\times\{\frac{1}{2}\}$ and also prove that the set of discontinuities of $f$ is: \begin{equation}\notag \left([1,0]\times [0,\frac{1}{2})\right)\cup \left([0,1]\times(\frac{1}{2},1]\right) \end{equation}
I have no idea how to start, also I would like to know if obtaining that result can help me to know if the iterated integrals of the function exist or not.
Welcome to math.stackexchange. Here's one approach using some topology. For showing continuity on $A:=[0,1]\times \{\frac{1}{2}\}$, it suffices to show that $(f|_A)^{-1}(V)$ is open (relative to $A$) for any open set $V\subset \mathbb{R}$. Let $V$ be an open set of $\mathbb{R}$. If $1\in V$, then $(f|_A)^{-1}(V)=[0,1]\times \{\frac{1}{2}\}$, which is open relative to $A$. Otherwise if $1\notin V$, then $(f|_A)^{-1}(V)=\varnothing$, which is also an open relative to $A$.