Let $\Omega$ be a measure space, a $\sigma$-algebra of subsets of $\Omega$ is said to be finite if it is generated by finitely many subsets $X_1,\dots,X_n\subseteq \Omega$.
Observation. A random variable $X$ on $\Omega$ is simple if and only if it is measurable w.r.t. a finite $\sigma$-algebra.
Proof. Suppose that $X=\sum _{i=1}^nx_i I_{A_i}$, then the $\sigma$-algebra induced by $X$ is precisely $\sigma(A_1,\dots,A_n)$, which is finite. Conversely, suppose that $X$ is measurable on $\Sigma=\sigma(B_1,\dots,B_n)$, then every element in $\sigma(B_1,\dots,B_n)$ is a finite union of elements in some measurable partition $\mathcal C=\{C_1,\dots,C_M\}$. Moreover, since $X^{-1}(B)$ is $\Sigma$-measurable for every Borel set $B$, the disjoint collection $\{X^{-1}(\{a\})\}_{a\in \R}$ must have at most finitely many ($\leq 2^M$) elements, thus we are done.
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