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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?

Akira
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1 Answers1

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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.

Mason
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