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Let $(X,\mathcal{A})$ and $(X',\mathcal{A}')$ be two measurable spaces. A function $f: X \to Y$ is called $\mathcal{A}$-$\mathcal{A}'$-measurable if $f^{-1}(\mathcal{A}') \subseteq \mathcal{A}$.

When it comes to work, i.e. if we consider real or extended real-valued functions, we most of the time use that a function is measurable if for all $\alpha \in \mathbb{R}$ one and therefore all of the following sets belong in $\mathcal{A}$: $$\{f > \alpha\} \qquad \{f \geq \alpha\}\qquad \{f < \alpha\}\qquad \{f \leq \alpha\}$$ This is $\mathcal{A}$-$\mathcal{B}$-measurability where $\mathcal{B}$ denotes the Borel $\sigma$-algebra on $\mathbb{R}$ (or $\overline{\mathbb{R}}$). Why do we equip $\mathbb{R}$ with the Borel $\sigma$-algebra and for example not with the Lebesgue $\sigma$-algebra $\mathcal{L}$? Is it because that $\mathcal{B}$ is sufficiently large for our purposes and $\mathcal{L}$ would be too restrictive?

TheGeekGreek
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If you demand $\mathcal{A}$-$\mathcal{L}$ measurability then you will have fewer measurable functions.

For instance, if you take $\mathcal{A} = \mathcal{L}$ as well, there are continuous functions which are not $\mathcal{L}$-$\mathcal{L}$ measurable (nor $\mathcal{B}$-$\mathcal{L}$ measurable). See Example of a continuous function that is not measurable for an example.

Nate Eldredge
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