Are probability spaces generally assumed to be complete measure spaces?

Question

In the study of conditional expectation in $L^2(\Omega, \mathcal A, \mathbb P)$, I came accross the following statement : "If $\mathcal B$ is a sub-$\sigma$-algebra of $\mathcal A$ then the sub-vector space $L^2(\Omega, \mathcal B, \mathbb P)$ of $L^2(\Omega, \mathcal A, \mathbb P)$ consisting of the $\mathcal B$-measurable random-variables of $L^2(\Omega, \mathcal A, \mathbb P)$ is closed".

One way to prove this involves the fact that if a sequence converges in $L^2(\Omega, \mathcal A, \mathbb P)$ then there is a sub-sequence converging almost surely.

The next step is the one that troubles me. I know that the everywhere pointwise limit of a sequence of $\mathcal B$-measurable functions is $\mathcal B$-measurable. My understanding is that in the more general case of an almost everywhere limit, some further assumption is required, such as completeness of the measure space. Unless $(\Omega, \mathcal A, \mathbb P)$ is complete, I'm not confident that this a.e. limit of a given sequence of $\mathcal B$-measurable random variables should have a $\mathcal B$-measurable representative in its equivalence class in the space $\mathcal L^2(\Omega, \mathcal A, \mathbb P)$.

Hence my question, are all probability spaces assumed to be complete ?

Answer 1

If $f_n \to f$ almost everywhere define $g(x)=\lim f_n(x)$ if the limit exist and $0$ otherwise. Then we can show that $g$ is measurable and $f=g$ a.e.. Hence $f$ and $g$ are equal as elements of $L^{2}$. Completeness of the measure space is not required for this argument.

Answer 2

Let's be more careful here about the space of functions ($\mathscr{L}^2$) and the space of equivalence classes of functions ($L^2$).

Fix a sequence $X_n\in\mathscr{L}^2(\Omega,\mathcal{B},\mathbb{P})$ of random variables that converges in the seminormed space $\mathscr{L}^2(\Omega,\mathcal{A},\mathbb{P})$ to some $X$. Then by extracting a subsequence, we may assume $X_n\to X$ $\mathbb{P}$-almost surely. Now the event "$X_n(\omega)$ is a Cauchy sequence" is in $\mathcal{B}$: it is $$ E=\bigcap_{m=1}^\infty\bigcup_{n=1}^\infty\bigcap_{k\geq n} [(X_k-X_n)^{-1}(B(0,m^{-1}))]. $$ and we know it has null complement. So we can construct $$ Y=\left[\omega\mapsto\begin{cases} \lim X_n(\omega) &\omega\in E\\ 0 & \omega\notin E \end{cases} \right] $$ (the $0$ is just a convention) and this $Y$ is a $\mathcal{B}$-measurable random variable that agrees with $X$ $\mathbb{P}$-a.s., so it is in $\mathscr{L}^0(\Omega,\mathcal{B},\mathbb{P})\cap\mathscr{L}^2(\Omega,\mathcal{A},\mathbb{P})=\mathscr{L}^2(\Omega,\mathcal{B},\mathbb{P})$. Finally, take quotient to get $L^2(\Omega,\mathcal{B},\mathbb{P})$.