Describe explicitly the $M$-measurable functions in case $M$ is one of the following $\sigma$-algebras:

(a) $M=\{\emptyset,X\}$

(b) $M=2^{X}$

(c) For certain disjoint sets $E_1,...,E_N$, $X=\cup_{k=1}^N E_k$, and $M$ is the algebra (in fact, $\sigma$-algebra) generated by the collection of sets $\{E_1,...,E_N\}$.

Here's my book's definition of $M$-measurable:

Suppose $f:X\to[-\infty,\infty]$. Then $f$ is $M$-measurable if for all $t\in[-\infty,\infty]$ the set $f^{-1}([-\infty,t])$ belongs to $M$. Inn other words, $\{x\in X|f(x)\le t\}\in M$.

Of course, the form of the inequality $f(x)\le t$ is arbitrary.



(a) One can prove that if $f\colon X\to [-\infty,\infty]$ is $\mathcal M$-measurable, then for all $t\in[-\infty,\infty]$, $f^{-1}([-\infty,t))$ is $\mathcal M$-measurable hence so is $f^{-1}\{t\}$. We have that $X=\bigsqcup_{t\in [-\infty,\infty]}f^{-1}\{t\}$ hence $f^{—1}\{t\}=X$ for a unique $t$.

(b) All the functions are measurable, as each subset is measurable.

(c) $\mathcal M$ is generated by the finite partition $E_1,\dots, E_N$. If $f=\sum_{j=1}^Na_jE_j$, then $f$ is $\mathcal M$-measurable. The converse also holds.

  • $\begingroup$ Is there any restriction on $a_j$? Thanks. $\endgroup$ – user70520 Sep 28 '13 at 18:22
  • $\begingroup$ I don't think (there are no problem of definition as the $E_j$ are disjoint). $\endgroup$ – Davide Giraudo Sep 28 '13 at 19:51
  • $\begingroup$ Wouldn't $E_{N+1}$ be just the empty set? $\endgroup$ – user70520 Sep 29 '13 at 0:32
  • $\begingroup$ I didn't realize that the $E_j$ cover $X$. I've edited accordingly. $\endgroup$ – Davide Giraudo Sep 29 '13 at 7:40

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