Let $A$ be a compact set of $\mathbb{R}^n$ and $f$ continuous on $A$. Let $F=f_A$, and let $I_0$ be a cube containing $A$. Divide $I_0$ in $2^n$ equal subcubes $I_{1_1},\dots, I_{1_{2^n}}$.On $I_0$ we define $F_0=max f_A$. We define $A_{k_i}=A\cap cl(I_{k_i})$ and $F_{1} (\mathbf{x})= \begin{cases} max f_{A_{1_i}}(x), & \text{if }x\in I_{1_i}\text{ and } A_{1_i}\text{ not empty}\\ 0, & \text{ otherwise} \end{cases}$.
Now, we repeat the same division of subcubes in subsubcubes, and so on, and define, for each new division, $F_{k} (\mathbf{x})= \begin{cases} max f_{A_{k_i}}(x), & \text{if }x\in I_{k_i}\text{ and } A_{k_i}\text{ not empty}\\ 0, & \text{ otherwise} \end{cases}$.
The exercise asks us to prove that $F_k$ converges uniformly to $f_A$ if and only if $f(\mathbf{x})=\mathbf{0} \ \ \forall \mathbf{x}\in fr(A)$
Exercise 10, section 5.5, Wendell Fleming's Functions of Several Variables.
My first doubt is: Shouldn't the $F_k$ be defined in terms of $ cl(I_{k_i})$ instead of just $I_{k_i}$?Because for a given point in $A$, there will be a certain order where that point will belong to $fr(I_{k_i})$, where $F_k(\mathbf{x})=0 $
My second doubt is how do I solve this? I have no clue as to how to tackle this. A hint would be just fine. Thanks.
