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I am trying to prove the next:

Let $\boldsymbol{C} = \{[n, n + 1) : n\in\mathbb{Z}\}.$ The $\sigma$-algebra generated by $\boldsymbol{C}$, $\sigma(\boldsymbol{C})$, is the collection of all countable unions of members of $\boldsymbol{C}$. An extended real-valued function defined on $\mathbb{R}$ is $\sigma(\boldsymbol{C})$-measurable if and only if it is a right-continuous step function with jump discontinuity occurring at integers in $\mathbb{R}$ only.

I have troubles with the part "if and only if": If the function is right-continuous step function with jump discontinuity is measurable because is the limit of linear combination of step function over subsets of the class $\boldsymbol{C}.$ For the other implication it seems "intuitively" true but I cannot see why; disjointness of the class could be the fact of $f$ has jump discontinuity at integers and be a step function but I am stuck in this.

Any kind of help is thanked in advanced.

Edit: The definition of measurability that I am following is:

Let $(X,\mathcal{A})$ be an arbitrary measurable space and let $D\in \mathcal{A}.$ An extended real-valued function f defined on $D$ is said to be $\mathcal{A}$-measurable on $D$ if it satisfies the condition that $\{x\in D: f(x)\leq \alpha\}\in\mathcal{A}$, that is, $f^{-1}([-\infty,\alpha])\in\mathcal{A}$, for every $\alpha\in\mathbb{R}.$

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  • $\begingroup$ Following a comment from Robert W. on my answer, can you precise on which $\sigma$-algebras are equipped the domain and codomain of your functions? $\endgroup$ – mathcounterexamples.net Sep 16 at 16:42
  • $\begingroup$ I've put the definition of measurability that I'm using. $\endgroup$ – Suiz96 Sep 16 at 17:04
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The result seems to be wrong. Consider the map $f$ defined as $$f(x)=\begin{cases} 1/4 & x \le 0\\ 1/4 + x/2 & 0 < x \le 1\\ 3/4 & 1 < x \end{cases}$$

$f$ is not a step function. However for $X \in \sigma(\boldsymbol{C})$, $f^{-1}(X) = \mathbb R$ if $[0,1) \subseteq X$ and $f^{-1}(X) = \emptyset$ otherwise. Proving that $f$ is $\sigma(\boldsymbol{C})$ measurable.

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  • $\begingroup$ Many thanks! That was a very nice counterexample. By the way, Your website is awesome!! $\endgroup$ – Suiz96 Sep 16 at 16:27
  • $\begingroup$ Thanks for the compliment. Really appreciated! $\endgroup$ – mathcounterexamples.net Sep 16 at 16:29
  • $\begingroup$ It seems that the question is about $\sigma(\boldsymbol{C})/\mathcal{B}(\mathbb{R})$ measurability. $\endgroup$ – Robert W. Sep 16 at 16:40

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