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Let $(f_n)_{n \geq 1}$ be a sequence of measurable functions from $(\Omega, \mathcal F)$ to a metric space $(X, d)$. Suppose that $f$ is such that

$$ \lim_{n \rightarrow +\infty} d(f_n(\omega),f(\omega)) = 0, \ \forall \omega \in \Omega. $$

Is the function $f$ $\mathcal F$-measurable?

What I know: if the space $X$ is separable (not necessarily complete) then this is true. I wonder why separability is needed here because I found an attempt that does not use separability. However, I still have a little bit of doubt so I would like to ask where did I go wrong...

My attempt: Let $C$ be a closed set then the map $x \mapsto d(x,C)$ is continuous. Call this map $\Psi_C$ then the composition maps $ \Psi_C \circ f_n$ is measurable. $\Psi_C \circ f_n \rightarrow \Psi_C \circ f$ pointwise so $\Psi_C \circ f$ is measurable too. Now I observe

$$ f^{-1}(C)= \{ \omega \in \Omega: d(f(\omega),C) = 0 \} = (\Psi_C \circ f)^{-1} \{ 0 \} \in \mathcal F $$

for every closed set $C$, which generates the Borel sets $\mathcal B(X)$ induced by $d$, and so $f$ is measurable.

A similar argument is given here. However, I don't understand why in most probability/ analysis books, they usually assume the metric space is separable to have the limit measurable.

Thanks in advance!

Jeffrey Jao
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    Your argument is fine. Separability is not needed for measurability of the limit. – geetha290krm Feb 23 '24 at 04:41
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    there is a notion of being $\mu$-measurable where $\mu$ is a measure. Then, to be $\mu$-measurable where $f$ is valued in a Banach space you need that the domain of $f$ would be $\mu$-almost separable valued. Then there is a convergence defined in this sense that is not equivalent to say that the final function is measurable. By example take a look at Analysis II of Herbert Amann and Joachim Escher – Masacroso Feb 23 '24 at 07:59

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