Let $X$ and $Y$ be topological spaces. By a simple function $\phi: X\to Y$ we mean a finite range Borel measurable function.

Q. Is the point-wise limit of a sequence of simple functions a Borel measurable function?

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    $\begingroup$ This question is better suited for MSE, in fact it´s answered here $\endgroup$
    – Saúl RM
    Jan 17, 2022 at 14:24
  • $\begingroup$ @Saúl Rodríguez Martín, It is concerned with set-valued functions not complex valued! I have no clue to find a proof $\endgroup$
    – ABB
    Jan 17, 2022 at 14:38
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    $\begingroup$ Your notation is unknown to me. What is the definition of a finite range Borel measurable function? $\endgroup$ Jan 17, 2022 at 15:15
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    $\begingroup$ In fact the standard definition is actually: simple function=finite range, Borel measurable function=linear combination of characteristic functions of Borel subsets. But where are the set-valued functions? $\endgroup$ Jan 17, 2022 at 21:07
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    $\begingroup$ @PietroMajer I was confused at first too, but I think he means that the function takes values in any set $Y$, not just in $\mathbb{C}$ (although that´s not what people usually mean when they say set-valued). $\endgroup$
    – Saúl RM
    Jan 18, 2022 at 22:16

1 Answer 1


The answer to your question is Yes provided the topology of $Y$ is such that for each non-empty open set $O\subset Y$ there is a strictly increasing sequence $(O_k)$ of open sets: $$ \overline O_k\subset O_{k+1}\subset O,\quad k=1,2,\ldots, $$ with $\bigcup_kO_k=O$. For suppose $(f_n)$ is a sequence of Borel functions from $X$ to $Y$ with pointwise limit $f$. With $O$ and the $O_k$ as above, $$ f^{-1}(O)=\bigcup_{k=1}^\infty\bigcup_{n=1}^\infty\bigcap_{m=n}^\infty f_m^{-1}(O_k). $$ This shows that $f^{-1}(O)$ is a measurable subset of $X$. Because the Borel $\sigma$-field on $Y$ is generated by the open sets, it follows that $f$ is Borel measurable.

In particular, the condition above is true if $Y$ is metrizable.

  • $\begingroup$ I do not think that you need a strictly increasing sequence with strict inclusions. And in fact that "strict" need not hold in metrizable spaces (e.g. if $Y$ is finite). $\endgroup$ Jan 18, 2022 at 17:37
  • $\begingroup$ I tried to indicate what I meant by strictly in the display just below that word. In particular, my use of $A\subset B$ follows the convention allowing for $A$ and $B$ to coincide. $\endgroup$ Jan 18, 2022 at 18:36

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