Lemma for Fubini’s Theorem

This post is about theorem 2.36 (stated below) in Folland’s book. This theorem is used as the main lemma for proving Fubini’s theorem. Here, I write about a lemma implicitly used in theorem 2.36. So, I guess a more appropriate title would be “Lemma for a Lemma for an Important Theorem.”

What About Theorem 2.36 Needs This Extra Attention?

Theorem 2.36: Suppose $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$ are $\sigma$-finite measure spaces. If $E\in\mathcal{M}\otimes \mathcal{N},$ then the functions $x\mapsto \nu(Ex)$ and $y\mapsto \mu(E^y)$ are measurable on $X$ and $Y$, respectively, and

$$\mu\times \nu(E)=\int\nu(E_x)d\mu(x)=\int \mu(E^y)d\nu(y)$$

Folland’s Idea: The theorem is first proved for finite spaces and then extended to $\sigma$-finite spaces. The first part is easy and so is the second part but it uses a lemma implicitly that might not be very obvious (at least to me when I first read it).To quote Folland:

Finally, if $\mu$ and $\nu$ are $\sigma$-finite, we can write $X\times Y$ as the union of an increasing sequence $\{X_j\times Y_j\}$ of rectangles of finite measure. If $E\in\mathcal{M}\otimes \mathcal{N},$ the preceding argument [referring to the proof for finite spaces] applies to $E\cap (X_j\times Y_j)$ for each $j$ to give

$$\mu\times \nu\big(E\cap(X_j\times Y_j)\big)=\int \chi_{X_j}\nu(E_x\cap Y_j)d\mu(x)=\int \chi_{Y_j}\nu(E^y\cap X_j)d\nu(y)$$

But here’s the catch. To apply the previous result, we need to consider how we define the spaces for our subproblem. Bear with me one moment. Let’s get some notational definitions out of the way first.

Definitions:

Let $X$ be any non-empty set. Consider any $\mathcal{E}\subset \mathcal{P}(X)$ and any non-empty $A\subset X.$ We define $\mathcal{E}\circ A$ as $$\mathcal{E}\circ A=\{E\cap A:E\in\mathcal{E}\}$$

We define $\mathcal{S}(X,\mathcal{E})$ as the $\sigma$-algebra generated by $\mathcal{E}$ on $X.$

The Hidden Issue: Coming back to our original problem, $E\cap (X_j\times Y_j)\in (\mathcal{M}\otimes \mathcal{N})\circ (X_j\times Y_j)$ but to apply the previous result we need to define the finite measure spaces $(X’,\mathcal{M}’,\mu’)$ and $(Y’,\mathcal{N}’,\nu’)$ for the subproblem and somehow make sure that $E\cap (X_j\times Y_j)\in \mathcal{M}’\otimes \mathcal{N}’.$

We define the spaces as $(X_j,\mathcal{M}\circ X_j,\mu|_{\mathcal{M}\circ X_j})$ and $(Y_j,\mathcal{N}\circ Y_j,\nu|_{\mathcal{N}\circ Y_j}).$

But do we know $E\cap (X_j\times Y_j)\in (\mathcal{M}\circ X_j)\otimes (\mathcal{N}\circ Y_j)$ for sure? All we know is $E\cap (X_j\times Y_j)\in (\mathcal{M}\otimes \mathcal{N})\circ (X_j\times Y_j).$ We need to prove that $(\mathcal{M}\circ X_j)\otimes (\mathcal{N}\circ Y_j)=(\mathcal{M}\otimes \mathcal{N})\circ (X_j\times Y_j).$

This wasn’t very trivial to me. It’s actually not difficult to prove. It’s just easy to miss.

Note that by proposition 1.3 (Folland), we need to show $$\mathcal{S}\big(X_j\times Y_j,\{P\times Q:P\in\mathcal{M}\circ X_j,Q\in \mathcal{N}\circ Y_j\}\big)=\mathcal{S}\big(X\times Y,\{P\times Q:P\in\mathcal{M},Q\in \mathcal{N}\}\big)\circ (X_j\times Y_j)$$

We slightly reframe this to an equivalent formulation $$\mathcal{S}\big(X_j\times Y_j,\{P\times Q:P\in\mathcal{M},Q\in \mathcal{N}\}\circ (X_j\times Y_j)\big)=\mathcal{S}\big(X\times Y,\{P\times Q:P\in\mathcal{M},Q\in \mathcal{N}\}\big)\circ (X_j\times Y_j)$$

Before proving this, I wish to note the following easy-to-prove technical lemma:

Lemma: If $\mathcal{M}$ is a $\sigma$-algebra on a non-empty set $X$ and $A\subset X$ is a non-empty subset then $\mathcal{M}\circ A$ is a $\sigma$-algebra on $A.$

Now, we can tackle the problem. As argued before, it reduces to proving the following theorem.

Theorem: For any non-empty set $X,$ any $\mathcal{E}\subset \mathcal{P}(X),$ and any non-empty subset $A\subset X,$ we have $\mathcal{S}(A,\mathcal{E}\circ A)=\mathcal{S}(X,\mathcal{E})\circ A$

Proof: Inclusion is obvious since $\mathcal{E}\subset \mathcal{S}(X,\mathcal{E})$ meaning $\mathcal{E}\circ A\subset \mathcal{S}(X,\mathcal{E})\circ A.$ Since $\mathcal{S}(X,\mathcal{E})\circ A$ is a $\sigma$-algebra containing $\mathcal{E}\circ A$, we must have $\mathcal{S}(A,\mathcal{E}\circ A)\subset\mathcal{S}(X,\mathcal{E})\circ A.$

For the other direction, let $\mathcal{M}=\{U\cup V:U\in\mathcal{S}(A,\mathcal{E}\circ A),V\in\mathcal{S}(A^c,\mathcal{E}\circ A^c)\}.$ Note that here complements are taken with respect to the parent set $X.$ We will first show that $\mathcal{M}$ is a $\sigma$-algebra on $X.$

The only non-trivial point is closure under complement. If $U\cup V\in \mathcal{M}$ then it is easy to show that $(U\cup V)^c=(A\setminus U)\cup(A^c\setminus V).$ Since $U\in\mathcal{S}(A,\mathcal{E}\circ A),$ a $\sigma$-algebra on $A,$ the complement of $U$ relative to $A$ must also be present in the said set. Hence, $(A\setminus U)\in \mathcal{S}(A,\mathcal{E}\circ A)$ and, similarly, $(A^c\setminus V)\in \mathcal{S}(A^c,\mathcal{E}\circ A^c).$ Thus, $(U\cup V)^c\in \mathcal{M}$ showing closure under complements.

For any $E\in \mathcal{E},$ we have $E=(E\cap A)\cup(E\cap A^c)$ meaning $E\in\mathcal{M}.$ Thus, $\mathcal{E}\subset \mathcal{M}.$ Since the later is a $\sigma$-algebra on $X$ we must have $\mathcal{S}(X,E)\subset \mathcal{M}.$ Hence $\mathcal{S}(X,E)\circ A\subset \mathcal{M}\circ A=\mathcal{S}(A,\mathcal{E}\circ A).$ Hence, proved.