Let $\Omega\subset \mathbb{R}^N$ be a open set. Consider $\Omega$ equipped with the Lebesgue measure and $\mathbb{N}$ equipped with the counting measure. Every function $g:\mathbb{N}\to\mathbb{R}$ is measurable with respect to the counting measure.

My question: Assume that $f:\Omega\times\mathbb{N}\to\mathbb{R}$ and for each fixed $i$, the function $f(\cdot,i)$ is measurable in $\Omega$. Can we conclude that $f$ is measurable?

Thank you


Define $f_n(x) = f(x, n)$. Let $V \subset \mathbb R$ be measurable. We have: $$ f^{-1}(V) = \bigcup_{n=1}^\infty f_n^{-1}(V) \times \{n\} $$

Each $f_n^{-1}(V)$ is measurable by hypothesis. $\{n\}$ is also measurable since $\mathbb N$ is equipped with the counting measure. Assuming that $\Omega \times \mathbb N$ is equipped with the product $\sigma$-algebra, $f_n^{-1}(V) \times \{n\}$ is measurable. The countable union is measurable too. Thus $f$ is measurable.

  • $\begingroup$ It seems that the fact that $f(x,\cdot)$ is measurable for $x$ fixed was useless. Am I right? $\endgroup$ – Tomás Sep 23 '13 at 20:19
  • $\begingroup$ This works if $ℝ$ is equipped with the Borel σ-Algebra, since it is generated by all open sets of $ℝ$. Does this also work if we equip $ℝ$ with the Lebesgue σ-Algebra obtained by completing the measure space $ℝ = (ℝ,\mathcal{F},λ)$? $\endgroup$ – k.stm Sep 23 '13 at 20:19
  • $\begingroup$ @K.Stm. I don't see why not. How do you define a measurable function? $\endgroup$ – Ayman Hourieh Sep 23 '13 at 20:21
  • $\begingroup$ This is strange. We can replace $\mathbb{N}$ by any countable space and we can even equip it with any measure that the argument still work? $\endgroup$ – Tomás Sep 23 '13 at 20:23
  • $\begingroup$ @Tomás I undeleted after rewriting the answer. Does it make sense now or am I still misunderstanding the problem statement? $\endgroup$ – Ayman Hourieh Sep 23 '13 at 20:37

Yes, because $$[f>a]=\cup ([f_i>a]\times\{i\}).$$


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