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Problem:

Suppose that $[a,b] \subset \mathbb{R}$, $(\Omega, \mathcal{F})$ is a measure space and $E$ is a topological space. Suppose $f : [a,b] \times \Omega \to E$ is such that:

  • $\forall t$ $f(t, \cdot): \Omega \to E$ is measurable. (Here the $\sigma$-fields of $E$ is the one generated by open sets)
  • $\forall \omega \in \Omega$ $f(\cdot, \omega) : [a,b] \to E$ is right-continuous.

How can I conclude that $f$ is $\mathcal{B}([a,b])\otimes\mathcal{F} - \mathcal{B}(E)$ measurable?

Attempt:

I tried in the following way. Let us define $f_n : [a,b] \times \Omega \to E$ in the following way, $f_n ([\frac{k}{n}, \frac{k+1}{n}), \omega)=f(\frac{k+1}{n}, \omega)$ for all $0 \leq k \leq n-1$. Since $f_n$ is measurable in $[\frac{k}{n}, \frac{k+1}{n}) \times \Omega$ bye the first hypothesis on $f$ we get that $f_n$ is measurable and right-continuous.

Now I noticed that $f_n$ point wise converge to $f$. In fact we have that for a fixed $(t,\omega)$ we obtain $t\in [\frac{k}{n}, \frac{k+1}{n})$ and thus $f_n(t,\omega)=f(\frac{k+1}{n},\omega) \to f(t,\omega)$ by the fact that $f$ is right-continuous.

Now I proved the following lemma:

Lemma: if $(X, \mathcal{X})$ is a measure space and $E$ is a metric space we have that if $f_n : X \to E$ are measurable and point wise converge to $f$ that $f$ is measurable.

Proof: If $A$ is an open set then: $$f^{-1}(A)=\bigcup\limits_{n} \bigcap\limits_{k \geq n} \{ x \in X: dist(f_k(x), A^c) > \frac 1 n \}$$ is a measurable set $\Box$.

Is this correct? Does this lemma hold also for general $E$ topological space? Are there different hypotheses to make on $E$?

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1 Answer 1

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The the argument of the OP works fine if $E$ is assumed to be a metric space.

The remaining of this posting is to show that the Lemma in the OP does not hold in general topological spaces (other than metric spaces)

Counter example(Dudley): Let $I$ be the unit interval in the real line, and equip $I^I$ with the topology of pointwise convergence (a.k.a. product topology). For each $n\in\mathbb{N}$ define the map $f_n:I\rightarrow I^I$ by letting $$f_n(x):y\mapsto \max(0,1−n|x−y|),\qquad x\in I$$

Each map $f_n : I \rightarrow I^I$ is continuous and thus measurable with the Borel structures on $I$ and $I^I$. The sequence $f_n$ converges pointwise to a limit $f$ , which maps every $x \in I$ to $1_{\{x\}} : y 􏰀\mapsto \mathbb{1}(x = y)$. The map $f$ is not Borel measurable! To see this, consider a nonmeasurable set $B\subset I$. The set $$U=\{\xi\in I^I :\exists x\in B \,\text{with}\, \xi(x)>0\}$$ is open in $I^I$ yet, $f^{−1}(U) = B$.

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