Question : Let $X$ be a set and $\mathcal{F}$ be a sigma algebra for $X$. Let $E_1$, ...., $E_k$ be subsets of $X$. If $g(x)$ = $\sum \alpha_j \chi_{E_j}$ is measurable (where $\alpha_j$'s are distinct), is each $E_j$ measurable?
My approach: If $E_j$'s are disjoint, then this is obvious as $E_j = g^{-1} (\{\alpha_j\})$. If not, then the idea is to may be split $E_j$'s into disjoint sets.
For that, we define for each subset $S$ of $\{1,2,3, .. ,k\}$, $A_S$ = set of all $x$ which belong to $E_s$ for each $s \in S$ but not in $E_t$ for $t \notin S$. It is clear that $A_S$ and $A_T$ are disjoint for distinct subsets $S$ and $T$ of$ \{1,2,3, .. ,k\}$. In addition, for each subset $S$ of $\{1,2,3, .. ,k\}$, define $\beta_S = \sum_{i \in S} \alpha_i$
Now we can check that $g(x) = \sum_S \beta_S \chi_{A_S}(x)$. This is because each $x$ contributes $\alpha_i$ to $g(x)$ if it is in $E_i$, otherwise $0$. Now all the $A_S$ are pairwise disjoint, but the $\beta_S$'s may not be distinct.
How can I proceed further?
