Step functions in $\Bbb R^n$

Question

Let $\phi(x,y)$ be a real and continuous function in $\mathbb{R^2}$ such that $\phi(0,0)=0$ and let $\alpha$ and $\beta$ be two step functions in $\mathbb{R^n}$. How can I prove that $\phi(\alpha,\beta)$ is a step function in $\mathbb{R^n}$?

Answer 1

Since $\alpha$ and $\beta$ are step functions, there exists $a_i,b_i \in \mathbb{R^n}$, $\alpha_{i} \in \mathbb{R}$ such that :

$$\alpha(x) = \sum_{i \in I} \mathcal{1}_{[a_i,b_i]}(x) \alpha_i$$ for all $x \in \mathbb{R}^n$ where $[a_i, b_{i}] \cap [a_j,b_j] = \emptyset$ for $j \neq i$ and $\cup_{i \in I} [a_i,b_i]=\mathbb{R}^n$ (the family of $[a_i,b_i]$ forms a partition of $\mathbb{R}^n$). The sum is taken over a countable set $I$. Using the same kind of notatations for $\beta$ yields :

$$\beta(x) = \sum_{i} \mathcal{1}_{[c_i,d_i]}(x) \beta_i$$

Now we want to evaluate $\phi((\alpha(x),\beta(x))$ for a fixed $x \in \mathbb{R}^n$. Using our notations we have :

$$\phi((\alpha(x),\beta(x))= \phi(\sum_{i} \mathcal{1}_{[a_i, b_i]}(x) \alpha_i ,\sum_{i} \mathcal{1}_{[c_i,d_{i}]}(x) \beta_i)$$ this expression simplifies a lot since there exists $i_0$ and $j_0$ such that $x \in [a_{i_0},b_{i_0}]$ and $x \in [c_{j_0},d_{j_0}]$, therefore we can write :

\begin{align*} \phi((\alpha(x),\beta(x)) & = \phi(\mathcal{1}_{[a_{i_0}, b_{i_0}]}(x) \alpha_{i_0},\mathcal{1}_{[c_{j_0},d_{j_0}]}(x) \beta_{j_0})\\ & = \phi(\alpha_{i_0},\beta_{j_0}) \end{align*} and this expression is the same for all $x \in [a_{i_0},b_{i_0}] \cap [c_{j_0},d_{j_0}]$, which can be written as a new interval $[e_i,f_i]$.

Finally, you have the decomposition :

$$\phi((\alpha(x),\beta(x))= \sum_{i} \phi(\alpha_{i_0},\beta_{j_0}) \mathcal{1}_{[e_i,f_i]}(x) $$ and you still have to prove that the new family $[e_i,f_i]$ is indeed a decomposition of $\mathbb{\mathbb{R}^n}$ which should be easy to conclude.