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In topology, the standard definition of continuity works in the "backward" direction, since it puts a condition on the pre-images of a function rather than images:

$f$ is continuous if the pre-image under $f$ of every open set is open.

However, there is an alternative definition of continuity that works in the "forward" direction:

$f$ is continuous if $p\in\overline A\implies f(p)\in\overline{f(A)}$.

The standard definition of measurable functions is analogous to the first definition of continuity (just replace "open" with "measurable"). Can measurable functions be defined without reference to pre-images, by analogy with continuity?

EDIT: The original version of this question did not specify that I was looking for a definition without any use of pre-images. The proposed duplicate has an answer that considers the "forward" direction, but it still uses pre-images in part of its formulation. First I made a new post with a refined question, but was then kindly instructed to re-open this one.

WillG
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    This is a very reasonable question, but I doubt there is a satisfying answer. When the adjective is "open", there is a related operator "closure" (the connection to "open" is that the closure is the complement of the interior of the complement). When the adjective is "measurable", we would need something like the complement of the ?? of the complement, where ?? would be (I guess) the largest measurable subset of $A$, which is not well-defined. – Greg Martin Jul 19 '21 at 16:41
  • @GregMartin The closure of $A$ is also the intersection of all closed sets containing $A.$ We could define the same operator for measurable sets, though the result will not be measurable in general. But it still might enable a definition of measurable functions. – WillG Jul 19 '21 at 16:48
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    There is a way to formulate "measurable space" as a particular type of uniform space. The measurable functions correspond to the uniformly continuous functions. And (I think?) a function $f$ is uniformly continuous iff it maps Cauchy nets to Cauchy nets. – GEdgar Jul 19 '21 at 20:05
  • @GEdgar Sounds interesting/promising. The epilogue of the answer in the proposed duplicate seemed to suggest a similar idea, but I don't know enough about filters/Stone spaces/Boolean algebras to understand it. It would be elegant if the measurable maps are precisely the continuous maps with respect to some topology. – WillG Jul 19 '21 at 20:10
  • @WillG: Images of measurable functions do not behave well. Part of the analysis of such things falls within the realm of Analytic sets. There is a whole branch in set theory that deals with these matters "Descriptive set theory". – Mittens Jul 21 '21 at 16:36
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    @OliverDiaz You could say the same thing of images of continuous functions in topology (open- and closed-ness is not preserved). However, there is still a good "forward" definition of continuity (see post)—it just doesn't rely on defining open/closed sets. Hence I wonder if the same could be true for measurability. – WillG Jul 21 '21 at 16:48
  • @WillG: But for continuity the characterization $f(\overline{B})\subset \overline{f(B)}$ is all you need. In measurability things become more involve. – Mittens Jul 21 '21 at 16:52
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    One possible route is via Lusin's theorem: a function is measurable if and only if, for every $\varepsilon > 0$, it is continuous except on a set of measure $\varepsilon$. – pseudocydonia Jul 30 '21 at 20:29
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    I happened to ask the same at https://mathoverflow.net/questions/453522/forward-definition-of-measurability , see there for an answer. – Moritz Schauer Aug 28 '23 at 15:23
  • The characterization at https://math.stackexchange.com/questions/285848/what-properties-are-preserved-under-a-measurable-mapping does not use preimages if you use nets instead of filters. – Eric Wofsey Oct 03 '23 at 02:40

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