1. ## Step Functions

Let $S = \displaystyle\bigcup_{r=1}^n E_n$ be the union of distinct intervals $E_1, \ldots E_n$ in $\mathbb{R}$ i.e. none of the intervals are the same

I am trying to show that $\chi_S$ is a step function $\implies E_1,\ldots E_n$ are disjoint. I am a bit stumped.

So assume $\chi_S = \lambda_1 \chi_{E_1} + \ldots + \lambda_n \chi_{E_n}$.

If there is only one non-empty intersection of two intervals, it is easy to show that this is impossible, but I am unable to prove it in the general case, i.e. when there is an arbitrary number of collections of intervals with non-empty intersection, has anyone got any tips? Thanks very much

2. Originally Posted by slevvio
Let $S = \displaystyle\bigcup_{r=1}^n E_n$ be the union of distinct intervals $E_1, \ldots E_n$ in $\mathbb{R}$ i.e. none of the intervals are the same

I am trying to show that $\chi_S$ is a step function $\implies E_1,\ldots E_n$ are disjoint. I am a bit stumped.

So assume $\chi_S = \lambda_1 \chi_{E_1} + \ldots + \lambda_n \chi_{E_n}$.

If there is only one non-empty intersection of two intervals, it is easy to show that this is impossible, but I am unable to prove it in the general case, i.e. when there is an arbitrary number of collections of intervals with non-empty intersection, has anyone got any tips? Thanks very much
I assume chi is the indicator function, allow me to use the bolded one. Assume that $\displaystyle \mathbf{1}_S=\sum_{j=1}^{m}\alpha_j \mathbf{1}_{A_j}$ where $\{A_j\}_{j\in[m]}$ is an interval partition of $S$. Assume that $\{E_k\}_{k\in[n]}$ is not a pairwise-disjoint set. Then, we may assume (with the possibility of relabeling) that $x\in E_1\cap E_2$. Show that this implies that $\mathbf{1}_S(x)\geqslant 2$.

If this was unhelpful, in the sense that I didn't really say anything substantiative, then just tell me and I'll try to give a better hint.

3. Well $\mathbf{1}_S (x)=1$ on $E_1, E_2,$ and $E_1 \cap E_2$. So $\mathbf{1}_S (x) = \lambda_1 + \lambda_2$ + other possible terms = $1$ on $E_1 \cap E_2$. I just don't see why this has to be $\ge 2$.

It is if $E_1$ and $E_2$ are the only sets that intersect, but if not you get random other sets coming and adding different values of positive or negative lambdas that alter the whole sum.