Is $f$ measurable?

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.
• It seems that the fact that $f(x,\cdot)$ is measurable for $x$ fixed was useless. Am I right? – Tomás Sep 23 '13 at 20:19
• 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 $ℝ = (ℝ,\mathcal{F},λ)$? – k.stm Sep 23 '13 at 20:19
• 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? – Tomás Sep 23 '13 at 20:23
Yes, because $$[f>a]=\cup ([f_i>a]\times\{i\}).$$