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Let $f_s: \mathbb R \to \mathbb R$ a be family of Borel measurable functions parameterized by $s\in \mathbb R$. Consider the limit function $$ F(t)=\limsup_{s\to 0} f_s(t). $$ Is the function $F$ Borel measurable. This seems to be not true in general.

Consider a locally finite Borel measure $\mu$ on $\mathbb R$. Is the function $$F(x)=\liminf _{r\to 0+} \frac{log (\mu([x-r, x+r]))}{\log r} $$ Borel measurable?

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    $\begingroup$ Yes, because you can take the $\liminf$ along the rationals. $\endgroup$
    – Christian Remling
    Commented Oct 8, 2017 at 3:43
  • $\begingroup$ @ Christian Remling: You are right. Thanks. $\endgroup$
    – ronggang
    Commented Oct 8, 2017 at 5:21
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    $\begingroup$ @ChristianRemling What if $t\mapsto f_s(t)$ is $1$ if $s$ is irrational and $0$ otherwise? Or is your remark only meant to apply to the example given? $\endgroup$
    – Michael Greinecker
    Commented Oct 8, 2017 at 9:11
  • $\begingroup$ @MichaelGreinecker: Just for this example, surely. $\endgroup$
    – Nate Eldredge
    Commented Oct 8, 2017 at 13:45
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    $\begingroup$ @MichaelGreinecker: Yes, of course the remark was about the "example" (if you want to call it that), which I thought was the actual question, since the OP answers the first part him/herself. I have made many silly mistakes on MO in the past, but I think I'd have to get quite a bit more senile still before I could think that $\limsup_{s\to 0} \ldots = \limsup_{s\to 0, s\in\mathbb Q} \ldots$ in general. $\endgroup$
    – Christian Remling
    Commented Oct 8, 2017 at 16:04

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The function $F$ need not be Borel, even if $(s,t) \mapsto f_s(t)$ is a Borel function on $\mathbb{R}^2$. I wrote down a counterexample on Math.SE.

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This will depend on how $F_s$ depends on $s$. If there is no restriction, then every function $F : \mathbb{R} \to \mathbb{R}_{>0}$ can be obtained that way:

Partition $\mathbb{R}$ into countable $R_s$ such that each $R_s$ has $0$ as accumulation point. Now let $F_s(x) = 0$ if $x \notin R_s$ and $F_s(x) = F(x)$ otherwise.

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