Is $f:[a,b] \times \Omega \to E$ measurable?

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 $$E$$?

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, $$f^{−1}(U) = B$$.