According to Lusin's theorem (and the slightly weaker converse of that result), measurable functions on locally compact topological spaces that are equipped with a Borel regular measures may be characterized as follows:

Definition. Let $X$ be a locally compact space and $\mu$ be a regular measure on $X$. We call $f:X\to\mathbb C$ measurable if for every $\varepsilon>0$ and for every compact set $K\subset X$, there exists a compact set $L\subset X$ such that $\mu(K\setminus L)<\varepsilon$, and $f$ restricted to $L$ is continuous.

In fact, Bourbaki defines a measurable functions as such in their volume on integration. To me, this definition of a measurable function seems much less intuitive than the one I am used to, where we require that the preimage of every measurable set be measurable. My question is therefore the following: Are there any situations in which this definition of measurability is clearly advantageous or easier to work with?

According to Lusin's theorem (and the slightly weaker converse of that result), measurable functions on locally compact topological spaces that are equipped with a regular measure may be characterized as follows:

Definition. Let $X$ be a locally compact space and $\mu$ be a regular measure on $X$. We call $f:X\to\mathbb C$ measurable if for every $\varepsilon>0$ and for every compact set $K\subset X$, there exists a compact set $L\subset X$ such that $\mu(K\setminus L)<\varepsilon$, and $f$ restricted to $L$ is continuous.

In fact, Bourbaki defines a measurable functions as such in their volume on integration. To me, this definition of a measurable function seems much less intuitive than the one I am used to, where we require that the preimage of every measurable set be measurable. My question is therefore the following: Are there any situations in which this definition of measurability is clearly advantageous or easier to work with?