# Why do we need hypothesis of complete measure in this version of Fubini's theorem?

I'm reading below Fubini's theorem in page 3 of this lecture note.

Let $$(X, \mathcal{A}, \mu)$$ and $$(Y, \mathcal{B}, \nu)$$ be complete measure spaces, let $$\gamma$$ be the product outer measure on $$X \times Y$$ constructed above, and suppose that $$f: X \times Y \rightarrow \mathbb{R}$$ is $$\gamma$$-integrable. Then

(i) $$f(x, y)$$ is a $$\mu$$-integrable function of $$x$$ for $$\nu$$-a.e. $$y \in Y$$;

(ii) $$\int_{X} f(x, y) d \mu(x)$$ is a $$\nu$$-integrable function of y;

(iii) $$\int_{Y}\left(\int_{X} f(x, y) d \mu(x)\right) d \nu(y)=\int_{X \times Y} f(x, y) d \gamma$$.

In the proof,

• $$C$$ is a $$\gamma$$-measurable set of finite measure.

• For all $$j$$, $$\left\{A_{i}^{j} \times B_{i}^{j}\mid i=1,2, \ldots\right\}$$ is pairwise disjoint family of $$\mathcal{A}, \mathcal{B}$$ rectangles.

• $$E=\cap_{j}\left(\cup_{i} A_{i}^{j} \times B_{i}^{j}\right) \setminus C$$.

In my understanding, $$E = \left [\cap_{j}\left(\cup_{i} A_{i}^{j} \times B_{i}^{j}\right) \right] \cap C^c$$ is $$\gamma$$-measurable. Hence the $$y$$-slice $$E_y$$ of $$E$$ defined by $$E_y \triangleq \{x \in X \mid (x, y) \in E\}$$ is also measurable by the lemma in this question. I mean by this lemma that we don't need the hypothesis of measure completeness to obtain the measurability of $$E_y$$.

However, the author said that

But $$\left.E \subset \cap\left(\cup_{i} E_{i}^{j} \times F_{i}^{j}\right)\right)$$ and $$\nu$$ is a complete measure, so the slice $$\{x:(x, y) \in E\}$$ is also in $$\mathcal{A}$$ and also has $$\mu$$-measure zero for $$\nu$$-a.e. $$y \in E$$.

So they mean the measure completeness is necessary for the slice $$E_y$$ to be measurable.

Could you please elaborate on my confusion?

It is because $$\gamma$$ is not the product measure $$\mu \times \nu$$. Rather, $$\gamma = \overline{\mu \times \nu}$$, that is, $$\gamma$$ is the completion of the product measure $$\mu \times \nu$$. The measure space you are working with is $$(X \times Y, \overline{\mathcal{A} \otimes \mathcal{B}}, \overline{\mu \times \nu})$$.
It is not always the case that $$E \in \overline{\mathcal{A} \otimes \mathcal{B}} \implies E_y \in \mathcal{A}$$. For example, take $$X = Y = [0, 1]$$ with Lebesgue measure $$m_1$$ and Lebesgue sigma-algebra $$L$$. Let $$S \subset [0, 1]$$ be a set which is not Lebesgue-measurable. Then $$E = S \times \{0\}$$ is Lebesgue-measurable with $$m_2(E) = 0$$ since $$S \times \{0\} \subset [0, 1] \times \{0\} \in L \otimes L$$ and $$m_2([0, 1] \times \{0\}) = 0$$. However, $$E_{0} = S$$ is not Lebesgue-measurable.