Why don't we always consider complete measure spaces?

Question

Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and let $X=(X_t)$ and $Y=(Y_t)$ two stochastic processes. I know for example that $X$ and $Y$ are indistinguishable if there is a set $N$ of measure $0$ s.t. for all $\omega \notin N$ we have $X_t=Y_t$ for all $t$, but we can't write $\mathbb P\{\forall t, \ X_t=Y_t\}$ since $\{\forall t, X_t=Y_t\}$ may be not $\mathcal F-$ measurable.

• The thing is if $Y$ is a copy of $X$ and $(\Omega ,\mathcal F,\mathbb P)$ is complete, then $\{\forall t,X_t=Y_t\}$ is $\mathcal F-$measurable.

• I also know that each measure space can be completed by adding sets of measure.

Questions :

So, why don't we always work with complete measure space (since they can be always completed), and avoid for example the problem of the measurability of $\{\forall t, X_t=Y_t\}$ if $Y$ is a copy of $X$ (or many other measurability problem) ?

In what working in a non complete measure space can be interesting, or at least more interesting than to work with it's completion ? (since a non complete measure space can always be completed).

Do you have an example where it's worth to work with the uncompleted measure space rather than with the completed space ?

Answer 1

The main advantage I can think of is that the composition of two $(\mathcal B,\mathcal B)$-measurable functions are also $(\mathcal B,\mathcal B)$-measurable whereas the composition of two $(\mathcal L,\mathcal B)$-measurable functions need not be $(\mathcal L,\mathcal B)$-measurable. Of course, here $\mathcal B$ is the family of Borel sets and $\mathcal L$ is the family of Lebesgue measurable sets.

Many analysis books mean to say $(\mathcal L,\mathcal B)$-measurable when they say measurable. This is why those books state that we need a continuous function $\varphi$ so that $\varphi\circ f$ is measurable, provided that $f$ is a measurable function.