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If $(X, \Sigma, \mu)$ is a complete measure space, and $f$ is a function that is defined almost everywhere, can I use the language that $f$ is measurable? What does it mean for this function that is defined everywhere except on a set of measure $0$ to be measurable? Does it just mean that it has a measurable extension?

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The following are equivalence:

  • $f$ have a measurable extension.

  • Preimage of any Borel set is measurable.

  • $f$ is measurable on the measure subspace containing the domain of $f$.

All of them can be proved by remembering that the domain of $f$ is a measurable set. And fill the part outside the domain with $0$ to obtain an extension.

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