Completeness of probability space is often equivalent to the sigma algebra of the space contains all $P$-null sets.
Why does this definition make intuitive sense? Why would it be "incomplete" if the space didn't contain all $P$-null sets?
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$\begingroup$We can find motivation for the definition of a complete measure by considering measures on product spaces. Let $(\mathbb R, \mathcal B, m)$ be the standard Lebesgue measure space on the real line. We would like to construct a two-dimensional Lebesgue measure $m^2$ on the plane $\mathbb R^2$ as a product measure. To do this, we must first choose a suitable $\sigma$-algebra to define $m^2$ on. Let $\mathcal B^2 = \mathcal B\otimes\mathcal B$ be the $\sigma$-algebra generated by the collection $\{A_1\times A_2: A_1,A_2\in\mathcal B\}$. Then $(\mathbb R^2,\mathcal B^2,m^2)$ is a measure space, but it has its problems. Since the Lebesgue measure of a countable set is zero, we have$$ m^2(\{0\}\times A) = m(\{0\})m(A)=0 $$for any $A\in\mathcal B$. However, if $A$ is a non-Lebesgue measurable subset of $\mathbb R$, then $m^2(\{0\}\times A$ is not defined, but$$ \{0\}\times A\subset\{0\}\times \mathbb R, $$and $m^2(\{0\}\times \mathbb R) = 0$. We would like for subsets of measurable sets to be measurable - for example, to make use of the inequality $\mu(A)\leqslant\mu(B)$ for a positive measure $\mu$ and $\mu$-measurable sets $A\subset B$.
So we construct the completion of $(\mathbb R^2,\mathcal B^2,m^2)$ as follows: let $Z = \{N\subset A: m(A) = 0\}$, $\overline{\mathcal B^2} = \sigma(Z,\mathcal B^2)$ and define the outer measure $\overline{m^2}:\overline{\mathcal B^2}\to[0,\infty)$ by$$ \overline{m^2}(A) = \inf\{m^2(B)\mid A\subset B\in\mathcal B^2\}. $$Then $(\mathbb R^2,\overline{\mathcal B^2},\overline{m^2})$ is a complete metric space.
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