In $\S 1.4$ of Real Analysis: Theory of Measure and Integration by J. Yeh, the author provided an example:

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

Now, it's easy to see that functions of such type are $\sigma(\boldsymbol{\mathfrak C}) / \boldsymbol{\mathfrak B}_\mathbb R$ measurable. But I am not able to formalize how it is the other way. Any hint would be appreciated.

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    $\begingroup$ Measurable functions are constant on atoms. $\endgroup$ Apr 23, 2023 at 12:20

1 Answer 1


If $f$ is $\sigma(\boldsymbol{\mathfrak C}) $-measurable then $f^{-1} (\{f(n)\})$ is a union of the (disjoint) intervals $[k,k+1)$ and it contains $n$. Hence it must contain $[n,n+1)$. Can you finish?

In general, measurable functions are constant in atoms.

  • $\begingroup$ Appreciate your quick comment and the answer. Now to consolidate my thoughts: $f$ is constant a.e. on each atom $[n, n+1) $ and if $[k, k+1) \in f^{-1} (\{f(n)\})\implies k = n$ and subsequently we get the required step function that is right continuous at each integer. Please correct me if I am wrong. $\endgroup$ Apr 23, 2023 at 16:16
  • $\begingroup$ Also, I would request you to add your comment in the answer. $\endgroup$ Apr 23, 2023 at 16:17
  • $\begingroup$ Your argument is correct. I have included my comment in the answer. @User1865345 $\endgroup$ Apr 23, 2023 at 23:01
  • $\begingroup$ Appreciate your advice. Thanks. $\endgroup$ Apr 24, 2023 at 2:03

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